#  Superconductivity-induced improper orders

András L. Szabó  Institute for Theoretical Physics, ETH Zurich, 8093 Zurich,
Switzerland  Aline Ramires  Paul Scherrer Institute, 5232 Villigen PSI,
Switzerland

###### Abstract

The study of improper phases in the context of multiferroic materials has a
long history, but superconductivity has yet to be connected to the network of
ferroic orders. In this work, we highlight an overlooked mechanism that
couples superconducting order parameters to odd-parity orders in the charge or
spin sectors such that the latter emerge as improper orders. For that, we
explore a novel perspective of nonsymmorphic symmetries based on extended
symmetry groups in real space. We highlight how nonsymmorphic symmetries can
generate rather nonintuitive couplings between order parameters. In
particular, we find that a bilinear in the superconducting order parameter can
couple linearly to odd-parity orders in centrosymmetric systems. Our findings
can account for the unusual phenomenology of CeRh2As2, a recently discovered
heavy fermion superconductor, and open the door for exploring nonsymmorphic
symmetries in the broader context of improper orders with potential
applications to functional materials.

The Landau theory of phase transitions has been a leading framework for
understanding ordered phases of matter. It has been successfully applied to
describe magnetic and electric ordering and their non-trivial interplay in
multiferroic systems, which are central in the pursuit of exotic functional
materials [1, 2]. Within multiferroics, improper ferroelectrics develop
electric polarization controlled by the development of a leading distortive or
magnetic order parameter [3, 4, 5, 6]. More generally, improper phases are
associated with order parameters that develop as a secondary effect of the
development of a leading order. The interplay of superconductivity with
magnetic and charge orders is empirically known in multiple families of
materials and extensively discussed in the framework of intertwined [7] or
vestigial orders [8]. Nevertheless, their relation in the context of improper
orders has not yet been fully investigated, as the complexity of
superconducting order parameters has only recently started to be acknowledged.
In this work, we explore the untapped realm of superconducting-induced
improper orders, highlighting the role of nonsymmorphic symmetry for the
development of unexpected couplings between order parameters, which can
potentially explain the unusual phenomenology recently reported in a material
in the family of heavy fermion systems.

The development of ordered phases of matter can be understood on the
phenomenological level based on the notion of symmetry breaking associated
with the onset of an order parameter. In crystalline solids, the primary
symmetries involved are spatial, generally accompanied by time-reversal
symmetry. Spatial symmetries include translations, rotations, and reflections,
as well as combinations of these. Particularly notable are nonsymmorphic
systems, as these feature symmetry transformations that are necessarily
accompanied by a fractional primitive lattice vector (PLV) translation. These
systems have been extensively explored in the context of topological band
structures [9, 10, 11, 12, 13], and such symmetries are key to protecting band
degeneracies of Bloch’s states with opposite parity at specific points in
momentum space [14, 15]. However, the effects of nonsymmorphic symmetries in
the context of ordered states of matter are less explored, and most efforts
have been focused on their topological classification [16, 17, 18]. Similarly,
much of the research in the context of superconductivity relies on the
analysis of point group symmetries, the crystalline symmetries that are left
once one factors out translations, a trivial procedure in symmorphic systems.
Nonsymmorphic symmetries complicate the process of factoring out translations,
making the group theoretical analysis and the classification of
superconducting states more cumbersome, particularly if one works in a
momentum space representation [15, 19]. Here, we take a complementary view of
nonsymmorphic symmetries in the context of ordered phases of matter. We
classify the order parameters in the superconducting, charge, and spin sectors
based on their textures directly in real space [20], taking explicit account
of the nonsymmorphic nature of the crystal [21]. From this analysis, we find
new types of coupling between superconducting order parameters and orders in
the spin or charge sectors, which can induce odd-parity improper orders by the
development of a primary order in the superconducting channel, see Fig. 1. Our
results could lead to new functionalities and technological applications by
highlighting an unapped mechanism for the development of improper orders and
bringing new connectivities between superconductivity and other functional
phases of matter.

Figure 1:  Order parameter coupling within Landau theory. Assuming a primary
superconducting order parameter $\mathbf{\Psi}$ (potentially multicomponent),
and a secondary order $\phi$, this figure highlights the distinct
phenomenology that emerges from two different types of coupling between them.
Left: For a biquadratic coupling,
$f_{\lambda}=\lambda|\phi|^{2}|\mathbf{\Psi}|^{2}$, there are two transition
temperatures and both order parameters onset with the characteristic
$\sqrt{T}$ temperature dependence. Right: for linear-quadratic coupling,
$f_{\lambda}\sim\lambda\phi(\Psi_{1}^{*}\Psi_{2}\pm\Psi_{1}\Psi_{2}^{*})$,
there is a single transition temperature and the subleading order parameter
$\phi$ onsets with a slower, linear temperaure dependence. Here $\lambda$ is a
generic coupling constant. More details are given in the SI.

The recently discovered heavy-fermion compound CeRh2As2, depicted in Fig. 2
(a), realizes an exotic phase diagram with two superconducting phases as a
function of a $c$-axis magnetic field [22]. While the pairing symmetry of the
two phases is to date unclear, nuclear magnetic resonance (NMR) [23] and
nuclear quadrupole resonance (NQR) [24] measurements on the As sites reveal
antiferromagnetic order coexisting with the low-, but not with the high-field
superconducting phase. Most intriguingly, specific heat and thermal expansion
measurements indicate a single phase transition [25], suggesting that the
onset of superconductivity coincides with that of magnetism in the entire low-
field phase, in a range of magnetic fields spanning 4 T. Both NMR and NQR
measurements observed site-selective broadening of the signal within the
magnetic phase: of the two inequivalent As sites in the unit cell, only one of
them experiences a change in the local magnetic field with the onset of
magnetic ordering. These observations constrain the possible textures for
magnetic moments localized at the Ce sites. Within homogeneous phases, the
only consistent order is a layered antiferromagnet, with magnetic moments
along the $c$-axis, changing sign across the sublattices [see Figs. 2 (b) and
3 (b)]. Two further families of solutions with in-plane magnetic moments can
be obtained by doubling the unit cell [see Figs. 2 (c) and 3 (c)]. Below, we
discuss how these constraints on the magnetic order impose strong restrictions
on the nature of the superconducting state in this system.

Figure 2: (a) Crystal structure of CeRh2As2, with centrosymmetric
nonsymmorphic space group $P4/nmm$ ($\\#$ 129). The Ce atoms span a body-
centered tetragonal lattice with the main fourfold rotation axis going through
these sites. The crystal field due to the Rh and As atoms (dark and light grey
spheres, respectively) breaks inversion symmetry at the Ce sites, generating a
Ce sublattice structure (blue and red spheres), characterizing this system as
locally noncentrosymmetric. Notably, the inversion center is located at the
midpoint between Ce sublattices, (b) Original unit cell (green) projected to
the $x-y$ plane containing two Ce atoms (1 and 2) and (c) enlarged unit cell
(magenta) containing four Ce atoms (1 through 4). Dashed line represents the
diagonal mirror plane, red cross is the global inversion centre, coinciding
with our chosen origin. Translation by one lattice constant (indicated by
$\mathbf{t_{y}}$) is a trivial operation in the original unit cell scenario,
as it takes a sublattice into itself. In contrast, in the enlarged unit cell
scenario a translation by one lattice constant constitutes a new operation.

Within the Landau formalism, the coexistence of superconducting and magnetic
orders implies a coupling between the corresponding order parameters
consistent with all operative symmetries. The most straightforward gauge-
invariant coupling preserving time-reversal symmetry is quadratic in both the
superconducting ($\Psi$) and magnetic ($M$) order parameters, e.g. $\sim
M^{2}|\Psi|^{2}$. This coupling, however, does not result in the same critical
temperatures for the two phases. From the phenomenology of improper orders, a
linear-quadratic coupling between $M$ and $\Psi$ with a dominant
superconducting order would lead to the onset of magnetism at the
superconducting critical temperature (see Fig. 1 and discussion in the SI).
For this type of coupling, we need to recourse to a multi-component
superconducting order parameter, $\boldsymbol{\Psi}=(\Psi_{1},\Psi_{2})$,
which would allow for a term $\sim
iM(\Psi_{1}\Psi_{2}^{\ast}-\Psi_{1}^{\ast}\Psi_{2})$ in the free energy.

In CeRh2As2, the homogeneous magnetic order is odd parity, as the magnetic
moments are opposite for sites related by inversion symmetry [see Fig. 3 (b)
and (c)]. Due to the global inversion symmetry, order parameters classified
from the perspective of point group symmetries are generally labeled as either
even or odd, and superconducting order parameter bilinears are invariably even
parity. This simplified view naively makes a linear-quadratic coupling between
magnetic and superconducting order parameters impossible in CeRh2As2, as it
would require $\Psi_{1}$ and $\Psi_{2}$ to be of opposite parity. Below, we
show that in the absence of accidental degeneracies, modulated superconducting
orders in nonsymmorphic systems can sustain multi-component order parameters
with opposite parity, allowing for this unusual coupling and for the emergence
of magnetism as an improper order triggered by a primary superconducting
order.

More concretely, taking as a working example the space group $P4/nmm$
($\\#129$), we now discuss how unusual irreducible representations (irreps)
with components of opposite parity emerge once we enlarge the symmetry group
by extending the unit cell to account for modulated order parameters. The 16
symmetry operations in $P4/nmm$ are generated by
$\bar{C}_{4z}=\\{C_{4z}|\mathbf{t_{x}}/2\\}$, a rotation by $\pi/2$ along the
z-axis followed by $\mathbf{t_{x}}/2$, half a PLV translation along the
x-axis; $\sigma_{d}=\\{\sigma_{d}|\mathbf{0}\\}$, a mirror reflection along
the diagonal plane ($x=y$); and inversion $i=\\{i|\mathbf{0}\\}$ (the complete
list of group operations in the standard Seitz notation is given in the SI).
Given the nonsymmorphic nature of the space group, these 16 symmetry
operations, when composed, do not close into themselves. If we are interested
in homogeneous phases, we can redefine the composition of these operations
modulo integer lattice vector translations such that they form a group. This
procedure corresponds to factoring out translations to determine the little
group at the $\Gamma$ point in momentum space [15], which is isomorphic to
$D_{4h}$. For this group, the symmetry elements are organized in 10 conjugacy
classes, leading to 10 irreps which are labeled as even or odd parity (see
details in the SI).

If we allow the system to develop modulations encompassing multiple unit cells
in a commensurate manner, we need to consider the corresponding wave-vector
dictating the modulation. If the wave vector corresponds to a point in
momentum space at the edge of the BZ, we expect unusual degeneracies in
nonsymmorphic systems, as is extensively discussed in the context of
electronic band structures in momentum space. Choosing the simplest scenario,
here we double the unit cell and introduce the notion of an _extended group_
by adding to the original group new symmetry operations, which take one
primitive unit cell into another in the doubled unit cell [20]; see Fig. 2 (b)
and (c). The extended symmetry group is formed by the original 16 operations
plus the composition of these with a PLV translation, here chosen to be
$E^{\prime}=\\{E|\mathbf{t_{y}}\\}$, which is the extension of the identity
operation $E=\\{E|\mathbf{0}\\}$ (extended symmetry operations are denoted
with a prime). Extending the group of symmetries to 32 elements leads to novel
irreps (see SI for details). If the space group is symmorphic, the total
number of irreps is simply doubled, and the new irreps behave as the original
ones up to an extra minus sign under operations including a PLV translation.
In contrast, if the group is nonsymmorphic, the irreps associated with the
extended group can be fundamentally distinct from the original irreps.

To understand how nontrivial irreps are generated, we rely on two elementary
results from group theory: (i) the number of irreps is equal to the number of
conjugacy classes; (ii) the dimensions of irreps should follow the identity
$\sum_{i}|d_{i}|^{2}=|\mathbf{G}|$, where $d_{i}$ is the dimension of the
$i$-th irreducible representation and $|\mathbf{G}|$ is the order of the group
(the number of elements in the group). The order of the extended group in our
example is twice the order of the original group. For symmorphic systems, the
number of irreps of a given dimension is also doubled, trivially satisfying
points (i) and (ii). For nonsymmorphic systems, the conjugacy classes in the
extended group are not twice as many as in the original group. This happens
because some of the original and extended operators necessarily coalesce into
a single conjugacy class. This point can be understood by considering the
conjugation of a generic spatial symmetry operation
$O_{B}=\\{R_{B}|\mathbf{t}_{B}\\}$ by another generic operation
$O_{A}=\\{R_{A}|\mathbf{t}_{A}\\}$:

$\displaystyle
O_{A}^{-1}.O_{B}.O_{A}=O_{C}=\\{R_{A}^{-1}R_{B}R_{A}|R_{A}^{-1}(R_{B}\mathbf{t}_{A}-\mathbf{t}_{A}+\mathbf{t}_{B})\\},$
(1)

where the dot denotes the composition of operations, and we used
$O_{A}^{-1}=\\{R_{A}^{-1}|-R_{A}^{-1}\mathbf{t}_{A}\\}$. The presence of
inversion operation in P4/nmm allows us to choose $O_{A}=\\{i|\mathbf{0}\\}$,
such that the RHS of the equation above reads $\\{R_{B}|-\mathbf{t}_{B}\\}$.
In the original symmetry group, associated with the primitive unit cell, if
$O_{B}$ is nonsymmorphic with $\mathbf{t}_{B}$ corresponding to half a PLV
translation, we can redefine
$-\mathbf{t}_{B}=\mathbf{t}_{B}+PLV\equiv\mathbf{t}_{B}$. As a consequence,
$O_{C}=O_{B}$ and we do not get any information about conjugacy classes. On
the other hand, in the extended symmetry group
$-\mathbf{t}_{B}=\mathbf{t}_{B}+PLV\not\equiv\mathbf{t}_{B}$, such that
$O_{C}=O_{B}^{\prime}\neq O_{B}$. The relation
$O_{A}^{-1}.O_{B}.O_{A}=O_{B}^{\prime}$ indicates that $O_{B}$ and
$O_{B}^{\prime}$ belong to the same conjugacy class in the extended group. We
conclude that all conjugacy classes containing nonsymmorphic operations in the
original symmetry group are enlarged in the extended symmetry group. This
coalescence of conjugacy classes tells us that we have less than twice as many
irreps in the extended group, and, consequently, the new (double-valued)
irreps are generally not one-dimensional. The complete analysis is summarized
in Table 1 and a more detailed discussion is provided in the SI. In this
example, there are 14 conjugacy classes, therefore 14 irreps. There are 4 new
two-dimensional irreps in the extended group (labelled as $E_{im}$,
$i=1,...,4$, with $m$ standing for mixed parity), with the unusual property of
having zero character associated with inversion symmetry.

$G_{16}^{9}$ | $E$ | $\tilde{C}_{2z}$ |  $2\bar{C}_{4z}$ |  $2\bar{\sigma}_{x}$ | $2{\sigma}_{d}$ |  $i$ |  $\tilde{\sigma}_{h}$ |  $2\bar{S}_{4}$ |  $2\bar{C}_{2x}$ | $2{C}_{2\bar{d}}$
---|---|---|---|---|---|---|---|---|---|---
| $\swarrow\searrow$ | $\swarrow\searrow$ |  $\Downarrow$ |  $\Downarrow$ | $\swarrow\searrow$ |  $\Downarrow$ |  $\Downarrow$ |  $\Downarrow$ |  $\Downarrow$ | $\swarrow\searrow$
$G_{32}^{2}$ |  $E$ |  $E^{\prime}$ |  $\tilde{C}_{2z}$ |  $\tilde{C}_{2z}^{\prime}$ |  $4\bar{C}_{4z}$ |  $4\bar{\sigma}_{x}$ |  $2{\sigma}_{d}$ |  $2{\sigma}_{d}^{\prime}$ |  $2i$ |  $2\sigma_{h}$ |  $4\bar{S}_{4}$ |  $4\bar{C}_{2x}$ |  $2{C}_{2\bar{d}}$ |  $2{C}_{2\bar{d}}^{\prime}$
$A_{1g}$ |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1
$A_{2g}$ |  1 |  1 |  1 |  1 |  1 |  -1 |  -1 |  -1 |  1 |  1 |  1 |  -1 |  -1 |  -1
$B_{1g}$ |  1 |  1 |  1 |  1 |  -1 |  1 |  -1 |  -1 |  1 |  1 |  -1 |  1 |  -1 |  -1
$B_{2g}$ |  1 |  1 |  1 |  1 |  -1 |  -1 |  1 |  1 |  1 |  1 |  -1 |  -1 |  1 |  1
$E_{g}$ |  2 |  2 |  -2 |  -2 |  0 |  0 |  0 |  0 |  2 |  -2 |  0 |  0 |  0 |  0
$A_{1u}$ |  1 |  1 |  1 |  1 |  1 |  -1 |  -1 |  -1 |  -1 |  -1 |  -1 |  1 |  1 |  1
$A_{2u}$ |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  -1 |  -1 |  -1 |  -1 |  -1 |  -1
$B_{1u}$ |  1 |  1 |  1 |  1 |  -1 |  -1 |  1 |  1 |  -1 |  -1 |  1 |  1 |  -1 |  -1
$B_{2u}$ |  1 |  1 |  1 |  1 |  -1 |  1 |  -1 |  -1 |  -1 |  -1 |  1 |  -1 |  1 |  1
$E_{u}$ |  2 |  2 |  -2 |  -2 |  0 |  0 |  0 |  0 |  -2 |  2 |  0 |  0 |  0 |  0
$E_{1m}$ |  2 |  -2 |  2 |  -2 |  0 |  0 |  2 |  -2 |  0 |  0 |  0 |  0 |  0 |  0
$E_{2m}$ |  2 |  -2 |  2 |  -2 |  0 |  0 |  -2 |  2 |  0 |  0 |  0 |  0 |  0 |  0
$E_{3m}$ |  2 |  -2 |  -2 |  2 |  0 |  0 |  0 |  0 |  0 |  0 |  0 |  0 |  2 |  -2
$E_{4m}$ |  2 |  -2 |  -2 |  2 |  0 |  0 |  0 |  0 |  0 |  0 |  0 |  0 |  -2 |  2
Table 1: Character table for P4/nmm modulo two integer lattice translations.
The first line gives the 10 conjugacy classes for P4/nmm modulo integer
lattice translations (isomorphic to $D_{4h}$ and the abstract group
$G_{16}^{9}$ [15]). The second line encodes the 14 conjugacy classes in P4/nmm
modulo two integer lattice translations (isomorphic to abstract group
$G_{32}^{2}$[15]). The symmetry operations are labeled according to their
associated point group operation. If the operation is accompanied by a half
PLV translation, it is marked with a bar, while if it is accompanied by two
orthogonal half PLV translations, it is marked with a tilde. Operations
without either a bar or a tilde are pure point operations. Operations marked
with a prime belong to the set of operations extended by a PLV translation.
The arrows indicate if the original conjugacy class splits
($\swarrow\searrow$) or doubles ($\Downarrow$) in the extended group. The
first 10 irreps have well-defined parity, and their labels follow that of the
$D_{4h}$ point group. The 4 last irreps are new to the extended group and have
mixed parity. The subscripts correspond to even (g), odd (u), or mixed (m)
parity. In gray color, we highlight the columns that are simply doubled for
the initial 10 irreps given the splitting on the conjugacy classes. In yellow,
we highlight the columns with zero characters due to the coalescence of the
original and extended operations in the same conjugacy class.

To better understand this last statement, we can conjugate inversion with a
nonsymmorphic operation. Choosing $O_{B}=\\{i|\mathbf{0}\\}$, we find the RHS
of Eq. 1 reads $\\{i|-2R_{A}^{-1}\mathbf{t}_{A}\\}$. If $O_{A}$ is
nonsymmorphic and $\mathbf{t}_{A}$ is half a PLV, $-2R_{A}^{-1}\mathbf{t}_{A}$
is a PLV, and therefore nonsymmorphic group elements inevitably connect
inversion $i=\\{i|\boldsymbol{0}\\}$ with $i^{\prime}=\\{i|\mathbf{t_{y}}\\}$,
leading to the enlargement of the associated conjugacy class under the group
extension. This result has strong implications. The fact that the counjugacy
class associated with the inversion operation is enlarged tells us that the
character associated with the new irreps is zero [15]. As a consequence, the
new two-dimensional irreps are associated with two-component basis functions
with components of opposite parity. This fact is directly associated with
known results in electronic band theory: our extended group is isomorphic to
the abstract group $G_{32}^{2}$, the little group at the $M$ point [15]. This
remarkable result challenges our intuition about inversion symmetry. In
centrosymmetric systems with nonsymmorphic symmetries inversion symmetry can
be “effectively broken” if the system develops textures with wave-vectors that
lie at the edge of the BZ.

Going back to the discussion on CeRh2As2, using the original group structure
modulo PLV (the little-group at the $\Gamma$ point), we can classify the order
parameters in the charge, spin, and superconducting sectors in terms of
irreps, which are either even or odd under parity (details in SI). In
contrast, if we classify the order parameters according to extended group (see
details in SI), we find orders associated with irreps labeled as $E_{im}$
having two components that transform differently under inversion. Curiously,
the magnetic order with in-plane moments discussed above belongs to the irrep
$E_{3m}$, but it cannot couple to any superconducting order through the
desired linear-quadratic term (see discussion in SI). On the other hand, the
magnetic order with moments along the $c$-axis is associated to the irrep
$A_{1u}$, which is odd parity, and can couple linearly to quadratic terms in
the superconducting order parameter if the latter is associated with irreps
$E_{3m}$ or $E_{4m}$. The latter type of coupling has strong implications for
the phenomenology. Magnetism develops as an improper order, and its
temperature dependence does not follow the standard $\propto\sqrt{T}$
behaviour expected for a leading order parameter within Landau theory, as
depicted in Fig. 1. Improper orders onset with a weaker temperature dependence
$\propto T$, which might make their experimental observation more difficult. A
slow onset for the improper magnetic order in CeRh2As2 is consistent with the
apparent magnetic critical temperature being lower than the superconducting
critical temperature [24, 23]. These results should trigger a more in-depth
investigation to determine the temperature at which magnetism emerges. If
magnetism onsets exactly at $T_{c}$, our analysis suggests that we have strong
constraints to the nature of both magnetism and superconductivity. It
necessarily indicates that the magnetic moments are aligned along the z-axis
and that superconductivity is of a very unusual type: a chiral PDW with two
components of opposite parity.

Figure 3: Representative orders. Inversion-odd $P_{z}$ polar (a) and $M_{z}$
layer antiferromagnet (b) orders, adhering to the original unit cell (green).
(c) Two components of $E_{3m}$ in-plane magnetic order (d) two components of
$E_{3m}$ superconducting order in the enlarged unit cell (magenta). The
pairing wave function is intrasublattice but antisymmetric between sites,
resulting in a modulation in the projected x-y plane ($+/-$, marked by green
and orange respectively). In (c) and (d) the components are related to each
other via 90 degree rotation, but one component is odd-parity, while the other
is even.

Classifying order parameters in real space taking into account
nonsymmorphicity by dealing with extended symmetry groups allows us to
systematically account for modulated unconventional orders and novel types of
couplings between them. Nonsymmorphic symmetries are related to intrinsically
complex crystalline structures. Nonsymmorphic crystals necessarily have
sublattice structures that add to the already complex set of charge, orbital,
and spin degrees of freedom necessary to faithfully describe the electronic
structure of most materials. The richness in the number of internal degrees of
freedom is nevertheless still strongly constrained by crystalline symmetries,
and the investigation of the development of improper orders can lead to very
refined information about the nature of order parameters developing in complex
materials. We expect that these findings can be harvested for a better
characterization of ordered phases of matter and in future studies of improper
orders of functional materials.

A.S. is grateful for financial support from the Swiss National Science
Foundation (SNSF) through Division II (No. 184739). A.R. also acknowledges
financial support from the Swiss National Science Foundation (SNSF) through an
Ambizione Grant No. 186043.

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Quadratic Equations in Elamite Mathematics

Nasser Heydari[Email<EMAIL_ADDRESS>and  Kazuo Muroi[Email<EMAIL_ADDRESS>
In this article, we study  some of quadratic equations and their solutions found  in the Susa Mathematical Texts (SMT). We show that the Susa scribes used this group of equations in different problems and took a  standard approach, known as completing the square, to find   solutions.

§ INTRODUCTION

Quadratic equations and their applications appear in several texts of the SMT. We  consider here  the quadratic equations  in SMT No. 5, SMT No. 6, SMT No. 20 and SMT No. 21. These texts were inscribed by Elamite scribes between 1894–1595 BC on   26 clay tablets excavated from Susa in  southwest Iran by French archaeologists in 1933. The texts of all the tablets,  along with their interpretations, were first published in 1961 (see [1]).

SMT No. 5[The reader can see   this tablet on the website of the Louvre's collection. Please see <https://collections.louvre.fr/ark:/53355/cl010185655> for obverse  and   reverse.]  is a collection of short problems which are intended to  teach    methods  for  solving  quadratic equationsquadratic equation. The text of this tablet consists of two parts: the mathematical part and the colophon part.  According to the colophon part,  the text originally contained 262 problems, of which   184 or more  are clearly legible. This tablet was written by two Susa scribesSusa scribes named qplDādi-šušnak (literally: a favorite of the city-godcity-god of Susa of Susacity-god of Susa) and qplSîn-išmeanni (literally: the moon-godmoon-god of Susa heard my request).

The content of  text of  SMT No. 6[The reader can see  this tablet on the website of the Louvre's collection. Please see <https://collections.louvre.fr/ark:/53355/cl010185656> for obverse  and   reverse.]   is similar to that of SMT No. 5, in that it contains  a collection of   short problems regarding quadratic equations  without   solutions. The first part deals with 39 quadratic equations  in the form of $x^2 + ax = b$, of which at least 32 problems remain. The second part of the text consists of 30 problems  in the form of  $x^2 - ax = b$, of which about 20 problems are comprehensible.

SMT No. 20[The reader can see   this tablet on the website of the Louvre's collection. Please see <https://collections.louvre.fr/ark:/53355/cl010186430> for obverse  and   reverse.]  contains two similar algebraic problems, one on the obverse and the other on the reverse of the tablet. Both problems solve quadratic equationsquadratic equation by completing the squarecompleting the square method. They  deal  with the geometric figure qplapusamikkumapusamikkum@apusamikkum (geometrical figure) that we have described elsewhere in  SMT No. 3 (see [5]).

The text  of SMT No. 21[The reader can see  this tablet on the website of the Louvre's collection. Please see <https://collections.louvre.fr/ark:/53355/cl010186431> for obverse  and   reverse.]   consists of two similar problems dealing with certain quadratic equationsquadratic equation concerning the areaarea of an apusamikkum of qplapusamikkumapusamikkum@apusamikkum (geometrical figure). The first problem is comparatively well preserved, but the second, which might have started on the obverse and continued on the reverse of the tablet, is mostly lost–leaving only the last steps of the solution.

§ QUADRATIC EQUATIONS

Quadratic equations belong to the category of polynomial equations. In such equations, one is usually interested in finding the zeros of a polynomial. Geometrically, these values are the $x$-coordinates of the intersection of the graph of the polynomial with the $x$-axis. Although finding the zeros of polynomials is not possible in many cases, for   polynomials of specific degrees this can be achieved.

§.§ General Form and Solutions

The general form of  quadratic equations is

\begin{equation}\label{eq-a}
\end{equation}

where $a\neq 0,b,c$ are real numbers and $x$ the unknown variable. The standard way of finding the real solutions of  a quadratic equation  is to use the quadratic formula

\begin{equation}\label{eq-b}
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},
\end{equation}

provided that the discriminant of the equation

$$\Delta = b^2-4ac$$

is non-negative. Note the if $\Delta <0$, one can consider the complex solutions of the equation.

The formula <ref> can be obtained by using an approach called completing the square. In this method, one can use the binomial identity

\begin{equation}\label{eq-c}
\end{equation}

to obtain formula <ref>.  If we multiply both sides of  <ref> with $a\neq 0$  and use $A=ax, B=\frac{b}{2}$, we can  write as follows:

\begin{align*}
&~~	a^2x^2+abx+ac   =0 \\
\Longrightarrow~~&~~ 	(ax)^2+2\left(\frac{b}{2}\right)(ax)+ac  =0 \\
\Longrightarrow~~&~~ 	(ax)^2+2\left(\frac{b}{2}\right)(ax)+\left(\frac{b}{2}\right)^2+ac =\left(\frac{b}{2}\right)^2 \\
\Longrightarrow~~&~~ 		\left(ax+\frac{b}{2}\right)^2  =\frac{b^2}{4}-ac \\
\Longrightarrow~~&~~ 			 ax+\frac{b}{2}  =\pm \sqrt{\frac{b^2}{4}-ac} \\
\Longrightarrow~~&~~ 			 ax  =- \frac{b}{2}\pm \frac{\sqrt{b^2-4ac}}{2} \\
\Longrightarrow~~&~~ 			 x  =  \frac{-b\pm \sqrt{b^2-4ac}}{2a}.
\end{align*}

In step 2, we added the term $\left(\frac{b}{2}\right)^2$ to both sides of the equation to complete the right-hand side of the binomial identity <ref> with $A=ax$ and $B=\frac{b}{2}$. This is the reason this method is usually called completing the square.

§.§ Quadratic System of Equations

Sometimes the sum and product of two quantities are known and we need to find the values of the quantities. For example, if the area and the perimeter of a rectangle are known, how can we determine its length and width. Such problems lead to solving  quadratic equations. Strictly speaking, let $x$ and $y$ be two unknown numbers such that

\begin{equation}\label{eq-ca}
\begin{cases}
\end{cases}
\end{equation}

where $p,q$ are known numbers. If we substitute the value of $y=\frac{q}{x}$ in the first equation and simplify, we get the following quadratic equation with respect to $x$:

\begin{equation}\label{eq-cb}
\end{equation}

By the quadratic formula, one can solve this equation to find the values of $x,y$:

\[ x=\frac{p+\sqrt{p^2-4q}}{2}~~~\text{and}~~~y=\frac{p-\sqrt{p^2-4q}}{2} \]

provided that $p^2-4q\geq 0$.

The  system of equations <ref>  belongs to  quadratic systems of equations because its solution requires  solving a quadratic equation.

Another way to solve <ref> is to use the algebraic identity

\begin{equation}\label{eq-ce}
\end{equation}

If we set $A=\frac{x}{2}$ and  $B=\frac{y}{2}$,  it follows from   <ref>   that

\begin{equation}\label{eq-cda}
\end{equation}

which consequently  gives us the following two equations

\begin{equation}\label{eq-cdb}
\frac{x+y}{2} =\sqrt{xy+\left(\frac{x-y}{2}\right)^2},
\end{equation}

\begin{equation}\label{eq-cdc}
\frac{x-y}{2} =\sqrt{\left(\frac{x+y}{2}\right)^2-xy}.
\end{equation}

it follows from <ref>, <ref> and <ref> that

\[ x=\ \frac{p+\sqrt{p^2-4q}}{2}~~~~~\text{and}~~~~~y=  \frac{p-\sqrt{p^2-4q}}{2},\]

\[ x=\ \frac{p-\sqrt{p^2-4q}}{2}~~~~~\text{and}~~~~~y=  \frac{p+\sqrt{p^2-4q}}{2}.\]

§.§ Geometric Viewpoint

In general, any function like $f(x)=ax^2+bx+c$ where $a\neq0,b,c$ are real numbers, is a quadratic function. The graph of such a function is a parabola in the $xy$-coordinate system. It can be shown that the graph of any quadratic function is obtained from the graph of the unit parabola $y=x^2$. This can be proved using completing the square as above. In fact,

\[ f(x)=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a} \]

This formula shows that the graph of $f(x)$ can be formed by translating, reflecting, compressing or stretching the  graph of $y=x^2$.

In this setting, the solutions of the quadratic equation $ ax^2+bx+c=0$ are exactly  the $x$-intercept points of the graph of $f(x)$. Depending on the discriminant of the equation, the quadratic function might have two, one or no $x$-intercept points (see <ref>).

Different parabolas

§.§ History

Quadratic equations appear in many Babylonian mathematical texts. It has been suggested that some  date to the Third dynasty of Ur  (circa 2000 BC). The Babylonian scribes were familiar with neither zero nor negative numbers, so the standard form of quadratic equations they usually considered have the following form:

\begin{equation}\label{equ-c}
ax^2\pm bx=c,
\end{equation}

where $a,b,c$ are non-zero positive numbers.  Based on mathematical texts in the Babylonian mathematical corpus, the algebraic method  Babylonian scribes used to solve the  quadratic equations  $ ax^2\pm bx=c$ was the standard method of completing the square. It should be noted that the Babylonian scribes only considered the positive solution for a quadratic equation.

In many   Babylonian texts, a quadratic system of equations like <ref> were treated. The Babylonian scribes used the method we explained in <ref> and  <ref> to solve such equations. Although in Babylonian mathematics   the  algebraic methods were used to complete the square,  geometric explanations for this method were  provided  later by   the Greek and Muslim mathematicians. Perhaps    Euclid might have been the one who  sparked other mathematicians' curiosity to tackle this issue, as he used a geometric argument  to show that (see [2])

“If a straight line is bisected and produced to any point, then the rectangle  contained by the whole (with the added straight line) together with the square  on half the line bisected is equal to the square  on the straight line made up of the half and the part added”.

We have pictured this situation in   <ref>. It is evident from       the figure that  the total area  of the colored rectangular regions on the left-hand side    is

\[x^2+\frac{ax}{2}+\frac{ax}{2}+\left(\frac{a}{2}\right)^2=x^2+ax+ \frac{a^2}{4}.\]

On the other hand, by cutting the orange rectangle  with sides $x$ and $\frac{a}{2}$  and attach it to  the empty rectangle  with the same dimensions, we get a square  of side $x+\frac{a}{2}$ whose area is exactly the same as the sum of areas of colored rectangles. In other words, we have

\[  x^2+ax+ \frac{a^2}{4}=\left(x+\frac{a}{2}\right)^2.\]

Euclid's completing the square

Mathematicians after Euclid also tried to use similar  geometric approaches   for solving quadratic equations. One of the greatest Muslim mathematicians   who gave a geometric treatment of this topic  was the prominent Persian mathematician    al-Khwarizmi  (circa 780–850).  He gave two different geometric solutions using completing the square  in order to solve the specific quadratic equation  $ x^2+10x=39$. We explain  his  two methods   in the following.

Al-Khwarizmi's First Method for Completing the Square

The geometric explanation of the  first method  is shown in    <ref>. Consider a rectangle with sides $x$ and $x+a$ whose area is the known value $b$ and then slice it off in order to produce  a square with side $x$ and a  rectangle  with sides $x$ and $a$. Next,  cut   off the obtained rectangle in half along the  dotted  red line to get two equal rectangles  with sides $x$ and  $\frac{a}{2}$. Then,  attach these two obtained rectangles along their sides   $x$ to  the top side and   the right  side of the square with side $x$. Finally,  add a small square  with side $\frac{a}{2} $ (and the area $ \frac{a^2}{4}$) to get a complete square  with side $x+\frac{a}{2}$. The area of the resulted square  is exactly the sum of our area   $ b $ and the extra area   $\frac{a^2}{4}$. In other words,

\[ \left(x+\frac{a}{2}\right)^2-\frac{a^2}{4}=x^2+ax. \]

Al-Khwarizmi's first method for completing the square

Al-Khwarizmi's Second Method for Completing the Square

In the second method, al-Khwarizmi carries out the procedure as  shown in    <ref>. Remove the rectangle  with sides $a$ and $x$ and cut it off along three  dotted red lines in order to get  four equal rectangles   with sides $\frac{a}{4}$ and $x $. 	Then, attach four obtained rectangles   along their sides $x$ to four sides of the square  with side $x$. Finally,   add four small squares   with sides $ \frac{a}{4}$ and areas $ \frac{a^2}{16}$ to get a complete square  with side $x+\frac{a}{2}$. The area of the resulted square is exactly the sum of our area $b$ and the extra four areas $\frac{a^2}{16} $. In other words,

\[ \left(x+\frac{a}{2}\right)^2-4\times\left(\frac{a^2}{16}\right)=x^2+ax. \]

Note that in both methods we have geometrically verified the algebraic identity

\[ \left(x+\frac{a}{2}\right)^2=x^2+ax+\frac{a^2}{4}. \]

Al-Khwarizmi's second method for  completing the square

As  Katz mentions in  [6],  Al-Khwarizmi's verbal descriptions of his  methods   are essentially the same as those of the Elamite and Babylonian scribes.   Geometric descriptions of his methods might  have been taken over from his  true  ancestors in the Near East!

§ QUADRATIC EQUATIONS IN SMT

§.§ SMT No. 5-Part 6:   $ ax^2=b$

In this subsection, simple quadratic equationsquadratic equation are presented without their solutions. The translations of two lines 1 and 2 are

(L1)  50,25 is the areaarea of a square (of a squaresquare). What is the side?

(L2)  $ \frac{1}{11}  $ of the areaarea of a square (of a squaresquare) is 4,35. What is the side?

§.§.§ Transliteration of Part 6

Obverse III: Lines 1-10

(L1) 50,25 a-šà nígin mi-nu

(L2) 11 a-šà 4,35 nígin mi-[nu]

(L3) 2 11 a-šà 9,10 nígin mi-n[u]

(L4) 1,41,40,25 a-šà nígin mi-nu

(L5)  11 11 a-šà 50,25 nígin mi-nu

(L6) 2 11 11 a-šà 1,40,50 nígin mi-[nu]

(L7) 41,10,25 a-šà nígin mi-n[u]

(L8) 11 7 a-šà 32,5 nígin mi-nu

(L9)  2 11 7 a-šà 1,4,10 nígin mi-nu

(L10)  2,44,41,40 a-šà nígin mi-nu

§.§.§ Mathematical Calculations of Part 6

(L1)  $ \displaystyle x^2=50,25$   and   $x=55$.

(L2)  $ \displaystyle \frac{1}{11}  x^2=4,35$   and   $x=55$.

(L3)  $ \displaystyle \left(2\times \frac{1}{11}\right) x^2=9,10$   and   $x=55$.

(L4)  $ \displaystyle x^2=1,41,40,50$   and   $x=10,5$.

(L5)  $ \displaystyle \left(\frac{1}{11}\times \frac{1}{11}\right) x^2=50,25$   and   $x=10,5$.

(L6)  $ \displaystyle \left(2\times \frac{1}{11}\times \frac{1}{11}\right) x^2=1,40,50$   and   $x=10,5$.

(L7)  $ \displaystyle x^2=41,10,25$   and   $x=6,25$.

(L8)  $ \displaystyle \left(\frac{1}{11}\times \frac{1}{7}\right) x^2=32,5$   and   $x=6,25$.

(L9)  $ \displaystyle \left(2\times \frac{1}{11}\times \frac{1}{7}\right) x^2=1,4,10$   and   $x=6,25$.

(L10)  $ \displaystyle x^2=2,44,41,40$   and   $x=12,50$.

§.§ SMT No. 5-Part 7:   $ (ax)^2=b$

In this subsection too, simple quadratic equationsquadratic equation are presented without their solutions, but the wording is ambiguous as the literal translation of line 19 shows

(L19) 35, the side of a squaresquare, the areaarea of a square, $ \frac{1}{7} $, what is the lengthlength?

With all data available, we cannot help judging that it means

\[``35^2 = \left(\frac{1}{7} x\right)^2,\  \mathrm{what\  is}\  x ?"\]

§.§.§ Transliteration of Part 7

Obverse III: Lines 11-30

(L11) 31(sic) nígin a-šà 2 uš nígin(sic) mi-[nu]

(L12) 31(sic) nígin a-šà 3 uš [mi-nu]

(L13) 30 nígin a-šà 4 [uš  mi-nu]

(L14) 30 nígin a-šà [$ \frac{\text{2}}{\text{3}} $ uš mi-nu]

(L15) 30 nígin a-šà [$ \frac{\text{1}}{\text{2}} $ uš mi-nu]

(L16) 30 nígin [a-šà $\frac{\text{1}}{\text{3}} $ uš mi-nu]

(L17) [30] nígin a-šà 4 uš [mi-nu]

(L18) [3]1(sic) nígin a-šà $\frac{\text{1}}{\text{3}}$ 4 uš mi-[nu]

(L19) 35 nígin a-šà 7 uš mi-nu

(L20) 35 nígin a-šà 2 7 uš mi-nu

(L21) 4,5 nígin a-šà 7 7 uš mi-nu

(L22) 4,5 nígin a-šà 2 7 7 uš mi-[nu]

(L23) 55 nígin a-šà 11 uš mi-[nu]

(L24) 55 nígin a-šà 2 11 uš mi-nu

(L25) 10,5 nígin a-šà 11 11 uš mi-nu

(L26) 10,5 nígin a-šà 2 11 11 uš mi-nu

(L27) 6,25 nígin a-šà 11 7 uš mi-nu

(L28) 6,25 nígin a-šà 2 11 8(sic) uš mi-nu

(L29) 12,50 nígin a-šà $ \frac{\text{2}}{\text{3}}~\frac{\text{1}}{\text{2}}~\frac{\text{1}}{\text{3}} $   11 7 uš mi-nu

(L30) 12,50 nígin 2(sic) a-šà $ \frac{\text{2}}{\text{3}}~\frac{\text{1}}{\text{2}}~\frac{\text{1}}{\text{3}}$  11 7 uš mi-nu

§.§.§ Mathematical Calculations of Part 7

(L11)  $ \displaystyle (2x)^2=30^2 $   and   $x=15$.

(L12)  $ \displaystyle (3x)^2=30^2 $   and   $x=10$.

(L13)  $ \displaystyle (4x)^2=30^2 $   and   $x=7;30$.

(L14)  $ \displaystyle \left(\frac{2}{3}x\right)^2=30^2 $   and   $x=45$.

(L15)  $ \displaystyle \left(\frac{1}{2}x\right)^2=30^2 $   and   $x=1,0$.

(L16)  $ \displaystyle \left(\frac{1}{3}x\right)^2=30^2 $   and   $x=1,30$.

(L17)  $ \displaystyle \left(\frac{1}{4}x\right)^2=30^2 $   and   $x=2,0$.

(L18)  $ \displaystyle \left[ \left(\frac{1}{3}\times \frac{1}{4}\right) x\right]^2=30^2 $   and   $x=6,0$.

(L19)  $ \displaystyle \left(\frac{1}{7}x\right)^2=35^2 $   and   $x=6,5$.

(L20)  $ \displaystyle \left[ \left(2\times\frac{1}{7}\right) x\right]^2=35^2 $   and   $x=3,2;30$.

(L21)  $ \displaystyle \left[ \left(\frac{1}{7}\times\frac{1}{7}\right)x\right]^2=(4,5)^2 $   and   $x=3,20,5$.

(L22)  $ \displaystyle \left[ \left(2\times\frac{1}{7}\times\frac{1}{7}\right) x\right]^2=(4,5)^2 $   and   $x=1,40,2;30$.

(L23)  $ \displaystyle \left( \frac{1}{11}  x\right)^2=55^2 $   and   $x=10,5$.

(L24)  $ \displaystyle \left[ \left(2\times\frac{1}{11} \right) x\right]^2=55^2 $   and   $x=5,2;30$.

(L25)  $ \displaystyle \left[ \left( \frac{1}{11}\times\frac{1}{11}\right) x\right]^2=(10,5)^2 $   and   $x=20,20,5$.

(L26)  $ \displaystyle \left[ \left(2\times\frac{1}{11}\times\frac{1}{11}\right) x\right]^2=(10,5)^2 $   and   $x=10,10,2;30$.

(L27)  $ \displaystyle \left[ \left(\frac{1}{11}\times\frac{1}{7}\right) x\right]^2=(6,25)^2 $   and   $x=8,14,5$.

(L28)  $ \displaystyle \left[ \left(2\times\frac{1}{11}\times\frac{1}{7}\right) x\right]^2=(6,25)^2 $   and   $x=4,7,2;30$.

(L29)  $ \displaystyle \left[ \left(\frac{2}{3} \times \frac{1}{2} \times \frac{1}{3}\times\frac{1}{11}\times\frac{1}{7}\right) x\right]^2=(12,50)^2 $   and   $x=2,28,13,30$.

(L30)  $ \displaystyle \left[ \left(2\times \frac{2}{3} \times \frac{1}{2} \times \frac{1}{3}\times\frac{1}{11}\times\frac{1}{7}\right) x\right]^2=(12,50)^2 $   and   $x=1,14,6,45$.

§.§ SMT No. 5-Part 8:   $ x^2+(ax)^2 =b$

In this subsection, simple quadratic equationsquadratic equation are    presented. The translation of line 31 is

(L31) The square (literally: areaarea of a square) of twice the lengthlength and the square (of the lengthlength) are added up to 1,15,0. (What is the lengthlength?)

§.§.§ Transliteration of Part 8

Obverse III: Lines 31-46

(L31) a-šà 2 uš ù a-šà gar-ma 1,15

(L32) a-šà 3 uš ù a-šà gar-gar-ma 2,[30]

(L33) a-šà 4 uš ù a-šà gar-gar-ma 4,1[5]

(L34) a-šà ù a-šà $ \frac{\text{2}}{\text{3}} $ uš gar-gar-ma 21,40

(L35) a-šà ù a-šà $ \frac{\text{1}}{\text{2}} $ uš gar-gar-ma 1[8],45

(L36) a-šà ù a-šà $\frac{\text{1}}{\text{3}} $ uš gar-gar-ma 16,40

(L37) a-šà ù a-šà 4 uš gar-gar-ma 15,56,15

(L38) a-šà ù a-šà $ \frac{\text{1}}{\text{3}} $ 4 uš gar-gar-ma 15,6,15

(L39) a-šà ù a-šà 7 uš gar-gar-ma 20,50

(L40) a-šà ù a-šà 2 7 uš gar-gar-ma 22,5

(L41) a-šà ù a-šà 7 7 uš gar-gar-ma 16,40,50

(L42) a-[šà ù a-šà] 2 7 7 uš gar-gar-ma 16,42,5

(L43) [a-šà] ù a-šà 11 uš gar-gar-ma [50],50

(L44) [a-šà ù a-šà] 2 11 uš gar-gar-ma 52,5

(L45) [a-šà ù a-šà 11 11 uš gar-gar-ma 1,4]1,40,50

(L46) [a-šà ù a-šà 2 11 11 uš gar-gar-ma 1,41,42,5]

§.§.§ Mathematical Calculations of Part 8

(L31)  $ \displaystyle  x^2+(2x)^2 =1,15,0 $   and   $x=30$.

(L32)  $ \displaystyle  x^2+(3x)^2 =2,30,0 $   and   $x=30$.

(L33)  $ \displaystyle x^2+(4x)^2 =4,15,0 $   and   $x=30$.

(L34)  $ \displaystyle x^2+\left(\frac{2}{3}x\right)^2=12,40 $   and   $x=30$.

(L35)  $ \displaystyle x^2+\left(\frac{1}{2}x\right)^2=18,45 $   and   $x=30$.

(L36)  $ \displaystyle x^2+\left(\frac{1}{3}x\right)^2=16,40 $   and   $x=30$.

(L37)  $ \displaystyle x^2+\left(\frac{1}{4}x\right)^2=15,56;15 $   and   $x=30$.

(L38)  $ \displaystyle x^2+\left[ \left(\frac{1}{3}\times\frac{1}{4}\right)x\right]^2=15,6;15 $   and   $x=30$.

(L39)  $ \displaystyle x^2+\left(\frac{1}{7}x\right)^2=20,50 $   and   $x=35$.

(L40)  $ \displaystyle x^2+\left[ \left(2\times \frac{1}{7}\right)x\right]^2=22,5 $   and   $x=35$.

(L41)  $ \displaystyle x^2+\left[ \left(\frac{1}{7}\times \frac{1}{7}\right)x\right]^2=16,40,5 $   and   $x=4,5$.

(L42)  $ \displaystyle x^2+\left[ \left(2\times \frac{1}{7}\times \frac{1}{7}\right)x\right]^2=16,42,5 $   and   $x=4,5$.

(L43)  $ \displaystyle x^2+\left( \frac{1}{11} x\right)^2=50,50 $   and   $x=55$.

(L44)  $ \displaystyle x^2+\left[ \left(2\times \frac{1}{11}\right)x\right]^2=52,5 $   and   $x=55$.

(L45)  $ \displaystyle x^2+\left[ \left(\frac{1}{11}\times \frac{1}{11}\right)x\right]^2=1,41,40,50 $   and   $x=10,5$.

(L46)  $ \displaystyle x^2+\left[ \left(2\times\frac{1}{11}\times \frac{1}{11}\right)x\right]^2=1,41,42,5 $   and   $x=10,5$.

§.§ SMT No. 5-Part 9:   $ x^2-(ax)^2 =b$

This subsection deals with quadratic equationsquadratic equation which are  conjugateconjugate quadratic equations to those in the previous subsection (see   <ref>). The statement of the problem is as follows:

(L7) The square (of a lengthlength) exceeds the square of $ \frac{1}{7} $ of the lengthlength by 20,0. (What is the lengthlength?)

§.§.§ Transliteration of Part 9

Reverse I: Lines 2-20

(L2) [a-šà ugu a-šà] $ \frac{\text{2}}{\text{3}} $ uš 8,20 dir[ig]

(L3) [a-šà ugu a-šà $ \frac{\text{1}}{\text{2}} $] uš 11,15 dirig

(L4) [a-šà ugu a-šà $ \frac{\text{1}}{\text{3}} $] uš 13,20 $<$dirig$>$

(L5) [a-šà ugu a-šà 4] uš 14,3,45 $<$dirig$>$

(L6) a-šà ugu a-šà [$ \frac{\text{1}}{\text{3}} $ 4] uš 14,53,45 $<$dirig$>$

(L7) a-šà ugu a-šà 7 uš 20 dirig

(L8) a-šà ugu a-šà 2 7 uš 18,45 $<$dirig$>$

(L9) a-š[à] ug[u] a-šà 7 7 uš 16,40 dirig

(L10) a-šà ugu a-šà 2 7 7 uš 16,42(sic),5(sic) $<$dirig$>$

(L11) [a-šà] ugu a-šà 11 uš 50 [dirig]

(L12) [a-šà] ugu a-šà 2 11 uš 48,4[5] [dir]ig

(L13) a-šà ugu a-šà 11 11 uš 1,41,40 [dirig]

(L14) a-šà ugu a-šà 2 11 11 uš 1,41,38,[4]5 $<$dirig$>$

(L15) a-šà ugu a-šà 11 7 uš 41,10 dirig

(L16) a-šà ugu a-šà 2 11 7 uš $<$4$>$1,8,45 dirig

(L17) a-šà ugu a-šà $ \frac{\text{2}}{\text{3}}~\frac{\text{1}}{\text{2}}~\frac{\text{1}}{\text{3}}$  11 7 uš 2,44,41,30

(L18)  8,45,55,33,20 dirig

(L19) a-šà ugu a-šà 2 $ \frac{\text{2}}{\text{3}}~\frac{\text{1}}{\text{2}}~\frac{\text{1}}{\text{3}} $ 11 7 uš 2,44,41,30

(L20)  5,3,42,13,20 dirig

§.§.§ Mathematical Calculations of Part 9

(L2)  $ \displaystyle x^2-\left(\frac{2}{3}x \right)^2 =8,20 $   and   $x=30$.

(L3)  $ \displaystyle x^2-\left(\frac{1}{2}x \right)^2 =11,15 $   and   $x=30$.

(L4)  $ \displaystyle x^2-\left(\frac{1}{3}x \right)^2 =13,20 $   and   $x=30$.

(L5)  $ \displaystyle x^2-\left(\frac{1}{4}x \right)^2 =14,3;45 $   and   $x=30$.

(L6)  $ \displaystyle x^2-\left[ \left(\frac{1}{3}\times \frac{1}{4}\right)x \right]^2 =14,53;45 $   and   $x=30$.

(L7)  $ \displaystyle x^2-\left(\frac{1}{7}x \right)^2 =20,0 $   and   $x=35$.

(L8)  $ \displaystyle x^2-\left[ \left(2\times \frac{1}{7}\right)x \right]^2 =18,45 $   and   $x=35$.

(L9)  $ \displaystyle x^2-\left[ \left(\frac{1}{7}\times \frac{1}{7}\right)x \right]^2 =16,40,0$   and   $x=4,5$.

(L10)  $ \displaystyle x^2-\left[ \left(2\times\frac{1}{7}\times \frac{1}{7}\right)x \right]^2 =16,38,45 $   and   $x=4,5$.

(L11)  $ \displaystyle x^2-\left(  \frac{1}{11}\ x \right)^2 =50,0 $   and   $x=55$.

(L12)  $ \displaystyle x^2-\left[ \left(2\times \frac{1}{11}\right)x \right]^2 =48,45 $   and   $x=55$.

(L13)  $ \displaystyle x^2-\left[ \left(\frac{1}{11}\times \frac{1}{11}\right)x \right]^2 =1,41,40,0$   and   $x=10,5$.

(L14)  $ \displaystyle x^2-\left[ \left(2\times\frac{1}{11}\times \frac{1}{11}\right)x \right]^2 =1,41,38,45 $   and   $x=10,5$.

(L15)  $ \displaystyle x^2-\left[ \left(\frac{1}{11}\times \frac{1}{7}\right)x \right]^2 =41,10,0 $   and   $x=6,25$.

(L16)  $ \displaystyle x^2-\left[ \left(2\times\frac{1}{11}\times \frac{1}{7}\right)x \right]^2 =41,8,45 $   and   $x=6,25$.

(L17-18)  $ \displaystyle x^2-\left[ \left(\frac{2}{3}\times\frac{1}{2}\times \frac{1}{3}\times\frac{1}{11}\times \frac{1}{7}\right)x \right]^2 =2,44,41,38;45,55,33,20 $   and

${}$  $x=12,50$.

(L19-20)  $ \displaystyle x^2-\left[ \left(2\times\frac{2}{3}\times\frac{1}{2}\times \frac{1}{3}\times\frac{1}{11}\times \frac{1}{7}\right)x \right]^2 =2,44,41,35;3,42,13,20 $   and

$ {}$ $x=12,50$.

§.§ SMT No. 5-Part 10:    $ x^2+ax =b$

In this subsection, a new group of  quadratic equations  are    presented which can be solved    by completing the square.
The translation of the first problem in line 21  is as follows:

(L21) To the square of my side one length  (my side) is added, and (the result is) 0;45. (What is my side?).

This means $x^2 + x = 0;45$  in    modern mathematical notation. In the following problems, the term “To the square of my side” is omitted.

It is   interesting   that the same quadratic equation  as $x^2 + x = 0;45$  occurs in the first problem of the mathematical tablet  BM 13901[This tablet  is one of the oldest known mathematical texts held in the British Museum. It contains about twenty-four problems and their solutions.]    which also deals with quadratic equations  and  uses completing the square  (see [7], for more information).

§.§.§ Transliteration of Part 10

Reverse I: Lines 21-47

(L21) a-na a-šà nigin-ia 1 uš dah-ma 45

(L22) 2 uš [da]h-ma 1,15

(L23) 3 uš [da]h-ma 1,45

(L24) 4 uš dah-ma 2,15

(L25) $ \frac{\text{2}}{\text{3}} $ uš dah-ma 35

(L26) $ \frac{\text{1}}{\text{2}} $ uš dah-ma 30

(L27) $ \frac{\text{1}}{\text{3}} $ uš dah-ma 25

(L28) 4 uš dah-ma 22,30

(L29) $ \frac{\text{1}}{\text{3}} $ 4 uš dah-ma 17,30

(L30) 7 uš dah-ma 25,2[5]

(L31) 2 7 uš dah-ma 30,2[5]

(L32) 7 7 uš dah-ma [16,45,25]

(L33) 2 7 7 uš dah-ma [16,50,25]

(L34) 1 $ \frac{\text{2}}{\text{3}} $ uš dah-ma [1,5]

(L35) 1 $ \frac{\text{1}}{\text{2}}$ uš dah-ma 1

(L36) 1 $ \frac{\text{1}}{\text{3}} $ uš dah-ma 55

(L37) 1 4 uš dah-ma 52,30

(L38) 1 $ \frac{\text{1}}{\text{3}} $ 4 uš dah-ma 47,30

(L39) 1 7 uš dah-ma 1,0,25

(L40) 1 2 7 uš dah-ma 1,5,25

(L41) 1 7 7 uš dah-ma 20,45(sic),25

(L42) 1 2 7 7 uš dah-ma 20,50(sic),25

(L43) 2 $ \frac{\text{1}}{\text{2}} $ uš dah-ma 1,30

(L44) 3 $ \frac{\text{1}}{\text{3}} $ uš dah-ma 1,55

(L45) 4 4 uš dah-ma 2,22,30

(L46) 7 igi-7 uš dah-ma 24,35

(L47) 7 2 igi-7 uš dah-ma 24,40

§.§.§ Mathematical Calculations of Part 10

(L21)  $ \displaystyle x^2 + x=0;45 $   and   $x=0;30$.

(L22)  $ \displaystyle x^2 + 2x=1;15 $   and   $x=0;30$.

(L23)  $ \displaystyle x^2 + 3x=1;45 $   and   $x=0;30$.

(L24)  $ \displaystyle x^2 + 4x=2;15 $   and   $x=0;30$.

(L25)  $ \displaystyle x^2 + \frac{2}{3}  x=0;35 $   and   $x=0;30$.

(L26)  $ \displaystyle x^2 + \frac{1}{2}  x=0;30 $   and   $x=0;30$.

(L27)  $ \displaystyle x^2 + \frac{1}{3}  x=0;25 $   and   $x=0;30$.

(L28)  $ \displaystyle x^2 + \frac{1}{4}  x=0;22,30 $   and   $x=0;30$.

(L29)  $ \displaystyle x^2 + \left( \frac{1}{3}\times \frac{1}{4} \right) x=0;17,30 $   and   $x=0;30$.

(L30)  $ \displaystyle x^2 + \frac{1}{7}  x=0;25,25 $   and   $x=0;35$.

(L31)  $ \displaystyle x^2 +\left( 2\times \frac{1}{7}\right)  x=0;30,25 $   and   $x=0;35$.

(L32)  $ \displaystyle x^2 +\left( \frac{1}{7}\times \frac{1}{7}\right)  x=16;45,25 $   and   $x=4;5$.

(L33)  $ \displaystyle x^2 +\left( 2\times \frac{1}{7}\times \frac{1}{7}\right)  x=16;50,25 $   and   $x=4;5$.

(L34)  $ \displaystyle x^2 +\left(1+ \frac{2}{3}\right)  x=1;5 $   and   $x=0;30$.

(L35)  $ \displaystyle x^2 +\left(1+ \frac{1}{2}\right)  x=1  $   and   $x=0;30$.

(L36)  $ \displaystyle x^2 +\left(1+ \frac{1}{3}\right)  x=0;55 $   and   $x=0;30$.

(L37)  $ \displaystyle x^2 +\left(1+ \frac{1}{4}\right)  x=0;52,30 $   and   $x=0;30$.

(L38)  $ \displaystyle x^2 +\left[1+ \left(\frac{1}{3}\times \frac{1}{4}\right)\right]  x=0;47,30 $   and   $x=0;30$.

(L39)  $ \displaystyle x^2 +\left(1+ \frac{1}{7}\right)  x=1;0,25 $   and   $x=0;35$.

(L40)  $ \displaystyle x^2 +\left[1+ \left(2\times\frac{1}{7}\right)\right] x=1;5,25 $   and   $x=0;35$.

(L41)  $ \displaystyle x^2 +\left[1+ \left(\frac{1}{7}\times\frac{1}{7}\right)\right]  x=20;50,25 $   and   $x=4;5$.

(L42)  $ \displaystyle x^2 +\left[1+ \left(2\times \frac{1}{7}\times\frac{1}{7}\right)\right]  x=20;55,25 $   and   $x=4;5$.

(L43)  $ \displaystyle x^2 +\left(2+ \frac{1}{2}\right)  x=1;30 $   and   $x=0;30$.

(L44)  $ \displaystyle x^2 +\left(3+ \frac{1}{3}\right)  x=1;55 $   and   $x=0;30$.

(L45)  $ \displaystyle x^2 +\left(4+ \frac{1}{4}\right)  x=2;22,30 $   and   $x=0;30$.

(L46)  $ \displaystyle x^2 +\left(7+ \frac{1}{7}\right)  x=24,35 $   and   $x=35$.

(L47)  $ \displaystyle x^2 +\left[7+ \left(2\times \frac{1}{7}\right)\right]  x=24,40 $   and   $x=35$.

§.§ SMT No. 5-Part 11-a:     $ x^2-ax =b$

This subsection contains quadratic equations which are conjugate  to   those in  the previous subsection (see   <ref>), although the first several lines are   lost.

Note that the scribes have implicitly assumed that $x>a$.  The statement of the first lost problem would be:

(L1)  From the square of my side one length  (my side) is subtracted, and (the result is) 14,30. (What is my side?)

This gives the quadratic equation  $x^2 - x = 14,30$  which appears  as the second problem in the mathematical tablet  BM 13901.

§.§.§ Transliteration of Part 11-a

Reverse II: Lines 1-17

(L1) [2 7 uš zi-ma 10,2]5

(L2) [7 7 uš zi-ma 1]6,35,25

(L3) [2 7 7 uš zi-m]a 16,30,25

(L4) [1 $ \frac{\text{2}}{\text{3}} $ uš z]i-ma 14,10

(L5) [1 $ \frac{\text{1}}{\text{2}} $ uš] zi-ma 14,15

(L6) [1 $ \frac{\text{1}}{\text{3}} $] uš zi-ma 14,20

(L7) [1] 4 uš zi-ma 14,22,30

(L8) 1 $ \frac{\text{1}}{\text{3}} $ 4 uš zi-ma 14,27,30

(L9) 1 7 uš zi-ma 19,45

(L10) 1 2 7 uš zi-ma 19,40

(L11) 1 7 7 uš zi-ma 12,35(sic),25

(L12) [1 2 7] 7 uš zi-ma 12,30(sic),25

(L13) [2 $ \frac{\text{1}}{\text{2}} $ uš] zi-ma 13,25(sic)

(L14) [3 $ \frac{\text{1}}{\text{3}}$ uš] zi-ma 13,20

(L15) [4 4 u]š zi-ma 12,52,30

(L16) [7 igi-7] uš zi-ma 16,15

(L17) [7 2 igi-7] uš zi-ma 16,10

§.§.§ Mathematical Calculations of Part 11-a

(L1)  $ \displaystyle x^2 -\left( 2\times \frac{1}{7} \right) x=0;10,25 $   and   $x=0;35$.

(L2)  $ \displaystyle x^2 -\left(\frac{1}{7} \times \frac{1}{7} \right) x=16;35,25 $   and   $x=4;5$.

(L3)  $ \displaystyle x^2 -\left(2\times \frac{1}{7} \times \frac{1}{7} \right) x=16;30,25 $   and   $x=4;5$.

(L4)  $ \displaystyle x^2 -\left( 1+ \frac{2}{3} \right) x=14,10 $   and   $x=30$.

(L5)  $ \displaystyle x^2 -\left( 1+ \frac{1}{2} \right) x=14,15 $   and   $x=30$.

(L6)  $ \displaystyle x^2 -\left( 1+ \frac{1}{3} \right) x=14,20 $   and   $x=30$.

(L7)  $ \displaystyle x^2 -\left( 1+ \frac{1}{4} \right) x=14,22;30 $   and   $x=30$.

(L8)  $ \displaystyle x^2 -\left( 1+ \frac{1}{3}+\frac{1}{4} \right) x=14,27;30 $   and   $x=30$.

(L9)  $ \displaystyle x^2 -\left( 1+  \frac{1}{7} \right) x=19,45 $   and   $x=35$.

(L10)  $ \displaystyle x^2 -\left[ 1+ \left(2\times \frac{1}{7}\right) \right] x=19,40 $   and   $x=35$.

(L11)  $ \displaystyle x^2 -\left[ 1+ \left(\frac{1}{7}\times\frac{1}{7}\right) \right] x=12;30,25 $   and   $x=4;5$.

(L12)  $ \displaystyle x^2 -\left[ 1+ \left(2\times \frac{1}{7}\times\frac{1}{7}\right) \right] x=12;25,25 $   and   $x=4;5$.

(L13)  $ \displaystyle x^2 -\left( 2+  \frac{1}{2} \right) x=13,45 $   and   $x=30$.

(L14)  $ \displaystyle x^2 -\left( 3+  \frac{1}{3} \right) x=13,20 $   and   $x=30$.

(L15)  $ \displaystyle x^2 -\left( 4+  \frac{1}{4} \right) x=12,52;30 $   and   $x=30$.

(L16)  $ \displaystyle x^2 -\left( 7+  \frac{1}{7} \right) x=16,15 $   and   $x=35$.

(L17)  $ \displaystyle x^2 -\left[ 7+ \left(2\times \frac{1}{7}\right) \right] x=16,10$   and   $x=35$.

§.§ SMT No. 5-Part 11-b:   $ ax-x^2  =b$

This short subsection completes the classification of quadratic equations. For example, line 19 says

(L19) Twice the side of a square  exceeds the area  by 0;45. (What is the side?)

§.§.§ Transliteration of Part 11-b

Reverse II: Lines 18-21

(L18) [1 nígin ugu] a-šà 15 dirig

(L19) 2 nígin ugu a-šà 45 dirig

(L20) $ \frac{\text{2}}{\text{3}} $ nígin ugu a-[š]à 5 dirig

(L21) $ \frac{\text{1}}{\text{2}} $ nígin ki-ma a-[šà]-ma

§.§.§ Mathematical Calculations of Part 11-b

(L18)  $ \displaystyle  x-x^2=0;15 $   and   $x=0;30$.

(L19)  $ \displaystyle  2x-x^2=0;45 $   and   $x=0;30$.

(L20)  $ \displaystyle  \frac{2}{3} x-x^2=0;5 $   and   $x=0;30$.

(L21)  $ \displaystyle \frac{1}{2}x=x^2  $   and   $x=0;30$.

§.§ SMT No. 6-Part 1:    $ x^2+ax =b$

§.§.§ Transliteration of Part 1

As a sample of the problems of this part, see the  Obverse II, lines 7-8 and   line 9 which contains the problem count.

(L7-8) To the square of my side $11+ \frac{2}{11}$ of the lengthlength (of my side) is added and (the result is) 1,0,40. What is the side?

(L9) (In total) 39 addition problems (in the form of $x^2 +ax = b$).

Obverse I: Lines 1-30

(L1) [a-šà nigin 1 uš dah 45] nigin mi-nu

(L2) [a-šà nigin 2 uš dah 1],15 nigin mi-nu

(L3) [a-šà nigin 3 uš dah]-ma 1,45 nigin mi-nu

(L4) [a-šà nigin 4] uš dah-ma 2,15 nigin mi-nu

(L5) [a-šà nigin $ \frac{\text{2}}{\text{3}} $] uš dah-ma 35 nigin mi-nu

(L6) [a-šà nigin $ \frac{\text{1}}{\text{2}} $] uš dah-ma 30 nigin mi-nu

(L7) [a-šà nigin] $ \frac{\text{1}}{\text{3}} $ uš dah-ma 25 nigin mi-nu

(L8) [a-šà nigin]-ia 4 uš dah-ma 22,30

(L9) [a-šà nigin]-ia $ \frac{\text{1}}{\text{3}} $ 4 uš dah-ma 17,30

(L10) [a-šà nig]in-ia 7 uš dah-ma 25,25

(L11) [a-šà nig]in-ia 2 7 uš dah-ma 30,25

(L12) [a-šà nigin-i]a 7 7 uš dah-ma 16,45,25

(L13) [a-šà nigin-i]a 2 7 7 uš dah-ma 16,50,25

(L14) [a-šà nigin 1]1 uš dah-ma 55,25

(L15) [a-šà nigin] 2 11 uš dah-ma 1,11(sic),25

(L16) [a-šà nigin] 11 11 [uš dah-ma 1,41,45,25]

(L17) [a-šà nigin] 2 11 11 uš dah 1,41,50,2[5]

(L18) [a-šà nig]in 11 7 uš dah 41,15,25

(L19) [a-šà nig]in 2 11 7 uš dah-ma 41,20,25

(L20) [a-šà] nigin $<$1$>$ $ \frac{\text{2}}{\text{3}} $ uš dah-ma 1,5

(L21) [a-šà nig]in 1 $ \frac{\text{1}}{\text{2}} $ uš dah-ma 1

(L22) [a-šà nig]in 1 $ \frac{\text{1}}{\text{3}} $ uš dah-ma 55

(L23) [a-šà nig]in 1 4 uš dah-ma 52,30

(L24) [a-šà nig]in 1 $ \frac{\text{1}}{\text{3}} $ 4 uš dah-ma 47,30

(L25) [a-šà] nigin 1 7 uš dah 1,0,25

(L26) [a-šà] nigin 1 2 7 uš dah 1,5,25

(L27) [a-šà] nigin 1 7 7 uš dah-ma 20,50,25

(L28) [a-šà] nigin 1 2 7 7 uš dah 20,55,25

(L29) [a-šà nig]in 1 11 uš dah [1,52,40],25

(L30) [a-š]à nigin [1 2 11 dah 1,53,35,25]

Obverse II: Lines 1-9

(L1) a-na a-šà nígin 7 igi-7 uš dah

(L2)    24,30(sic) nigin mi-[nu]

(L3) a-na a-šà nigin-ia 7 2 igi-7 uš

(L4)    dah-ma 24,35(sic) nigin mi-nu

(L5) a-na a-šà nigin-ia 11 $<$igi-11$>$ uš dah-ma

(L6)      1,0,35 nigin mi-nu

(L7) a-na a-šà nigin-ia 11 2 igi-11 $<$uš$>$ dah-ma

(L8)      1,0,40 nigin mi-nu

(L9) 39 dah

§.§.§ Mathematical Calculations of Part 1

Obverse I, Lines 1-30:

(L1)  $ \displaystyle   x^2+x=0;45 $   and   $x= 0;30$.

(L2)  $ \displaystyle   x^2+2x=1;15 $   and   $x= 0;30$.

(L3)  $ \displaystyle   x^2+3x=1;45 $   and   $x= 0;30$.

(L4)  $ \displaystyle   x^2+4x=2;15 $   and   $x= 0;30$.

(L5)  $ \displaystyle   x^2+\frac{2}{3}x=0;35 $   and   $x= 0;30$.

(L6)  $ \displaystyle   x^2+\frac{1}{2}x=0;30 $   and   $x= 0;30$.

(L7)  $ \displaystyle   x^2+\frac{1}{3}x=0;25 $   and   $x= 0;30$.

(L8)  $ \displaystyle   x^2+\frac{1}{4}x=0;22,30 $   and   $x= 0;30$.

(L9)  $ \displaystyle   x^2+\left(\frac{1}{3}\times\frac{1}{4}\right) x=0;17,30 $   and   $x= 0;30$.

(L10)  $ \displaystyle   x^2+\frac{1}{7}x=0;25,25 $   and   $x= 0;35$.

(L11)  $ \displaystyle   x^2+\left(2\times \frac{1}{7}\right) x=0;30,25 $   and   $x= 0;35$.

(L12)  $ \displaystyle   x^2+\left(\frac{1}{7}\times\frac{1}{7}\right) x=16;45,25 $   and   $x= 4;5$.

(L13)  $ \displaystyle   x^2+\left(2\times\frac{1}{7}\times\frac{1}{7}\right) x=16;50,25 $   and   $x= 4;5$.

(L14)  $ \displaystyle   x^2+ \frac{1}{11}  x=0;55,25 $   and   $x= 0;55$.

(L15)  $ \displaystyle   x^2+\left(2\times\frac{1}{11}  \right) x=1;0,25 $   and   $x= 0;55$.

(L16)  $ \displaystyle   x^2+\left(\frac{1}{11}\times\frac{1}{11}\right) x=1,41;45,25 $   and   $x= 10;5$.

(L17)  $ \displaystyle   x^2+\left(2\times\frac{1}{11}\times\frac{1}{11}\right) x=1,41;50,25 $   and   $x= 10;5$.

(L18)  $ \displaystyle   x^2+\left(\frac{1}{11}\times\frac{1}{7}\right) x=41;15,25 $   and   $x= 6;25$.

(L19)  $ \displaystyle   x^2+\left(2\times\frac{1}{11}\times\frac{1}{7}\right) x=41;20,25 $   and   $x= 6;25$.

(L20)  $ \displaystyle   x^2+\left(1+\frac{2}{3}\right) x=1;5 $   and   $x= 0;30$.

(L21)  $ \displaystyle   x^2+\left(1+\frac{1}{2}\right) x=1  $   and   $x= 0;30$.

(L22)  $ \displaystyle   x^2+\left(1+\frac{1}{3}\right) x=0;55 $   and   $x= 0;30$.

(L23)  $ \displaystyle   x^2+\left(1+\frac{1}{4}\right) x=0;52,30 $   and   $x= 0;30$.

(L24)  $ \displaystyle   x^2+\left[1+\left(\frac{1}{3}\times\frac{1}{4}\right)\right] x=0;47,30 $   and   $x= 0;30$.

(L25)  $ \displaystyle   x^2+\left(1+\frac{1}{7}\right) x=1;0,25 $   and   $x= 0;35$.

(L26)  $ \displaystyle   x^2+\left[1+\left(2\times\frac{1}{7}\right)\right] x=1;5,25 $   and   $x= 0;35$.

(L27)  $ \displaystyle   x^2+\left[1+\left(\frac{1}{7}\times\frac{1}{7}\right)\right] x=20;50,25 $   and   $x= 4;5$.

(L28)  $ \displaystyle   x^2+\left[1+\left(2\times\frac{1}{7}\times\frac{1}{7}\right)\right] x=20;55,25 $   and   $x= 4;5$.

(L29)  $ \displaystyle   x^2+\left(1+ \frac{1}{11}\right) x=1,52;40,25 $   and   $x= 10;5$.

(L30)  $ \displaystyle   x^2+\left[1+ \left(2\times \frac{1}{11}\right)\right] x=1,53;35,25 $   and   $x= 10;5$.

Obverse II, Lines 1-9:

(L1-2)  $ \displaystyle   x^2+\left(7+   \frac{1}{7}\right) x=24,35 $   and   $x= 35$.

(L3-4)  $ \displaystyle   x^2+\left[7+ \left(2\times \frac{1}{7}\right)\right] x=24,40 $   and   $x= 35$.

(L5,6)  $ \displaystyle   x^2+\left(11+   \frac{1}{11}\right) x=1,0,35 $   and   $x= 55$.

(L7-8)  $ \displaystyle   x^2+\left[11+  \left(2\times \frac{1}{11}\right)\right] x=1,0,40 $   and   $x= 55$.

§.§ SMT No. 6-Part 2:   $ x^2-ax =b$

The second part begins with the statement of a problem:

From the square of my side one lengthlength (of my side) is subtracted, and (the result is) 14,30. (What is my side?)

The last line contains the problem count:

(In total) 30 subtraction problems (in the form of $x^2 - ax = b$).

§.§.§ Transliteration of Part 2

Obverse II: Lines 10-27

(L10) i-na a-šà nígin-ia 1 uš zi-ma 1[4,30]

(L11) i-na a-šà nigin-ia $<$2$>$ uš zi-ma 1[4]

(L12) i-na a-šà nigin 3 uš zi-ma 13,30

(L13) i-na a-šà nigin 4 uš zi-ma 13

(L14) i-na a-šà nigin $ \frac{\text{2}}{\text{3}} $ uš zi-ma 14,[40]

(L15) i-na a-šà nigin $ \frac{\text{1}}{\text{2}} $ uš zi-ma 14,[4]5

(L16) [i]-na a-šà nigin $ \frac{\text{1}}{\text{3}} $ uš zi-ma [14,50]

(L17) [i]-na a-šà nigin 15 uš zi-ma 14,[5]2,30

(L18) [i-na] a-šà nigin $ \frac{\text{1}}{\text{3}} $ 4 uš zi 14,57,30

(L19) [i-na] a-šà nigin 7 uš zi-ma 15,24(sic)

(L20) [i-na] a-šà nigin-ia 2 7 uš zi-ma 10,25

(L21) [i-na] a-šà nigin 7 7 uš zi 16,35,25

(L22) i-na a-šà nigin 2 7 7 uš zi-[ma 1]6,30,25

(L23) i-na a-šà nigin 11 11 uš z[i-ma 1,41,35,2]5

(L24) i-na a-šà nigin 2 11 [1]1 uš [zi-ma 1,41,30,25]

(L25) i-na a-š[à nigin $\ldots$  $\ldots$ $\ldots$]

(L26) i-na a-[šà nigin $\ldots$ $\ldots$ $\ldots$ $\ldots$]

(L27) i-[na a-šà nigin $\ldots$ $\ldots$ $\ldots$]

Reverse I: Lines 1-14

(L1) i-[na a-šà nigin $\ldots$ $\ldots$ $\ldots$]

(L2)  i-na a-šà nigin 3 $ \frac{\text{1}}{\text{3}} $ [uš zi-ma]

(L3)   13,20

(L4)  i-na {1} a-šà nigin 4 uš 4 uš z[i-ma]

(L5)   12,57(sic),30

(L6)  i-na a-šà nigin-ia 7 igi-7 uš [zi-ma]

(L7)   16,15

(L8)  i-na [a]-šà nigin-ia 7 $<$2$>$ igi-7 $<$uš$>$ zi-ma

(L9)   16,10

(L10)  i-na a-šà nigin-ia 11 igi-11 zi-ma

(L11)   40,1[5]

(L12)  i-na [a-šà nigin 1]1 uš 2 ig[i-11 zi-ma]

(L13)   [40,10]

(L14)   30 [zi]

§.§.§ Mathematical Calculations of Part 2

Obverse II, Lines 10-27:

(L10)  $ \displaystyle   x^2-x=14,30 $   and   $x= 30$.

(L11)  $ \displaystyle   x^2-2x=14,0 $   and   $x= 30$.

(L12)  $ \displaystyle   x^2-3x=13,30 $   and   $x= 30$.

(L13)  $ \displaystyle   x^2-4x=13,0 $   and   $x= 30$.

(L14)  $ \displaystyle   x^2-\frac{2}{3} x=14,40 $   and   $x= 30$.

(L15)  $ \displaystyle   x^2-\frac{1}{2}x=14,45 $   and   $x= 30$.

(L16)  $ \displaystyle   x^2-\frac{1}{3}x=14,50 $   and   $x= 30$.

(L17)  $ \displaystyle   x^2-(0;15)x=14,52;30 $   and   $x= 30$.

(L18)  $ \displaystyle   x^2-\left(\frac{1}{3}\times\frac{1}{4} \right) x=14,57;30 $   and   $x= 30$.

(L19)  $ \displaystyle   x^2-\frac{1}{7}x=0;15,25 $   and   $x= 0;35$.

(L20)  $ \displaystyle   x^2-\left(2\times\frac{1}{7} \right)x=0;10,25 $   and   $x= 0;35$.

(L21)  $ \displaystyle   x^2-\left(\frac{1}{7}\times\frac{1}{7} \right)x=16;35,25 $   and   $x= 4;5$.

(L22)  $ \displaystyle   x^2-\left(2\times\frac{1}{7}\times\frac{1}{7} \right)x=16;30,25 $   and   $x= 4;5$.

(L23)  $ \displaystyle   x^2-\left(\frac{1}{11}\times\frac{1}{11} \right)x=1,41;35,25 $   and   $x= 10;5$.

(L24)  $ \displaystyle   x^2-\left(2\times\frac{1}{11}\times\frac{1}{11} \right)x=1,41;30,25 $   and   $x= 10;5$.

Reverse I, Lines 1-14:

(L2-3)  $ \displaystyle   x^2-\left(3+\frac{1}{3} \right)x=13,20 $   and   $x= 30$.

(L4-5)  $ \displaystyle   x^2-\left(4+\frac{1}{4} \right)x=12,52;30 $   and   $x= 30$.

(L6,7)  $ \displaystyle   x^2-\left(7+\frac{1}{7} \right)x=16,15 $   and   $x= 35$.

(L8-9)  $ \displaystyle   x^2-\left[7+\left(2\times\frac{1}{7}\right) \right]x=16,10 $   and   $x= 35$.

(L10-11)  $ \displaystyle   x^2-\left(11+\frac{1}{11} \right)x=40,15 $   and   $x= 55$.

(L12-13)  $ \displaystyle   x^2-\left[11+\left(2\times\frac{1}{11}\right) \right]x=40,10 $   and   $x= 55$.

§.§ SMT No. 20: First Problem

§.§.§ Transliteration

Obverse:  Lines 1-12

(L1) a-pu-sà-am-mi-ik-k[um a-šà ù uš ul-gar]

(L2) 36,40 [mi-nu uš]

(L3) za-e 36,40 gar 2[6,40 igi-gub]

(L4) a-na 36,40 i-ší-ma [16,17,46,40 ta-mar]

(L5) tu-úr-ma 1 uš 1/2 h[e-pe 30 ta-mar]

(L6) 30 nigin 15 ta-mar 1[5 a-na 16,17,46,40 dah]

(L7) 31,17,46,40 [ta-mar mi-na íb-si]

(L8) 43,20 íb-si 30 [ta-ki-il-ta]

(L9) i-na 43,20 zi-ma [13,20 ta-mar]

(L10) igi-26,40 igi-[gub pu-ţú-úr]

(L11) 2,15 ta-mar 2,1[5 a-na 13,20 i-ší-ma]

(L12) 30 ta-mar [30 uš]

§.§.§ Translation

Obverse:  Lines 1-12

(L1) The (figure called) qplapusamikkum. I added the area and the length (of a quadrant) together, (and the result is)lengthareaquadrantapusamikkum@apusamikkum (geometrical figure)

(L2) 0;36,40. What is the length?length

(L3) You, put down 0;36,40, (and) 0;26,40 of the (area) constant.area

(L4) Multiply (0;26,40) by 0;36,40, and you see 0;16,17,46,40.

(L5) Return. Halve 1 of the length, (and) you see 0;30.length

(L6) Square 0;30, (and) you see 0;15. Add 0;15 to 0;16,17,46,40, (and)

(L7) you see 0;31,17,46,40. What is the square root?square root

(L8) 0;43,20 is the square root. 0;30 (used in) completing the square,square root

(L9) subtract from 0;43,20, and you see 0;13,20.

(L10) Make the reciprocal of 0;26,24 of the (area) constant, (and)reciprocal of a numberarea

(L11) you see 2;15. Multiply 2;15 by 0;13,20, and

(L12) you see 0;30. 0;30 is the length.length
completing the square method

§.§.§ Mathematical Calculations

Consider the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) $ \Lambda$ as shown  in   <ref>.   If we denote the lengthlength of four equal quadrants of the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) by $x$, then, as we saw in lines 22-24 of SMT No. 3SMT No.c@SMT No. 3 (see [5]),  its areaarea of an apusamikkum and its diagonaldiagonal  can be obtained by

\begin{equation}\label{equ-SMT20-a}
S_{\Lambda}\approx (0;26,40)\times x^2
\end{equation}

\begin{equation}\label{equ-SMT20-b}
\overline{BD}\approx (1;20)\times x
\end{equation}

Diagonal of an  qplapusamikkum

By lines 1-2, we must have $S_{\Lambda}+\overline{BD}=0;36,40 $, so      <ref> and <ref>    give  us  the following quadratic equationquadratic equation:

\begin{equation}\label{equ-SMT20-c}
(0;26,40) x^2 + x = 0;36,40.
\end{equation}

The Susa scribeSusa scribes has used  completing the squarecompleting the square method to solve this quadratic equation. According to lines 3-4, he apparently uses the new variable

\begin{equation}\label{equ-SMT20-d}
y = (0;26,40)\times x
\end{equation}

and then multiplies both sides of <ref> by $0;26,40 $ to get a new quadratic equationquadratic equation with respect to new variable $y$:

\begin{equation*}
\left[(0;26,40)\times x\right]^2 + (0;26,40)\times x =(0;26,40)\times (0;36,40)
\end{equation*}

\begin{equation}\label{equ-SMT20-e}
y^2 + y =  0;16,17,46,40.
\end{equation}

Now,  according to lines 3-9

\begin{align*}
&~~  y^2 + y =  0;16,17,46,40 \\
\Longrightarrow~~&~~    y^2 + 2\times\left[(0;30)\times y\right] =  0;16,17,46,40 \\
\Longrightarrow~~&~~    y^2 + 2\times\left[(0;30)\times y\right] +(0;30)^2=  0;16,17,46,40+(0;30)^2 \\
\Longrightarrow~~&~~    y^2 + 2\times\left[(0;30)\times y\right] +(0;30)^2=  0;16,17,46,40+ 0;15  \\
\Longrightarrow~~&~~ (y+0;30)^2  =  0;31,17,46,40 \\
\Longrightarrow~~&~~ y+0;30=   \sqrt{0;31,17,46,40}\\
\Longrightarrow~~&~~ y+0;30=   \sqrt{(0;43,20)^2}\\
\Longrightarrow~~&~~  y+0;30= 0;43,20\\
\Longrightarrow~~&~~  y= 0;43,20-0;30
\end{align*}

which gives the solution of equation <ref> as

\begin{equation}\label{equ-SMT20-f}
\end{equation}

Finally,  according to lines 10-12, it follows  from <ref> and <ref> that

\[x= \dfrac{1}{(0;26,40)}  \times (0;13,20)=(2;15)\times (0;13,20)=0;30 \]

which is the solution of equation <ref>, namely the lengthlength of the “side” of the apusamikkumapusamikkum@apusamikkum (geometrical figure).

§.§ Second Problem

§.§.§ Transliteration

Reverse:  Lines 13-26

(L13) a-šà  uš  ù bar-d[á ul-gar 1,16,40]

(L14) za-e 26,40 igi-[gub šà ap-sà-mi-ki]

(L15) a-na 1,16,40 i-[ší 34,4,26,40 ta-mar]

(L16) {pu} tu-úr-ma [1 uš ù 1,20 bar-dá]

(L17) šà la-a ti-du-ú [ul-gar 2,20 ta-mar]

(L18) 1/2 2,20 he-pe 1,[10 ta-mar 1,10 nigin 1,21,40 ta-mar]

(L19) 1,21,40 a-na 34,[4,26,40 dah]

(L20) 1,55,44,[26,40 ta-mar]

(L21) mi-na íb-si 1,23,20 [íb-si]

(L22) 1,10 ta-[ki-il]-ta i-[na 1,23,20 zi-ma]

(L23) 13,2[0 ta-mar] igi-[26,40]

(L24) pu-ţú-[ú]r 2,15 ta-[mar]

(L25) 2,15 a-na 13,20 i-ší-ma

(L26) 30 ta-mar 30 uš

§.§.§ Translation

Reverse:  Lines 13-26

(L13) I added the area, the length, and the diagonal, (and the result is) 1;16,40.lengtharea

(L14) You, put down 0;26,40 of the (area) constant of qplapusamikkum,area of an apusamikkumapusamikkum@apusamikkum (geometrical figure)

(L15) multiply (it) by 1;16,40, (and) you see 0;34,4,26,49.

(L16) Return. 1 of the length and 1;20 of the diagonallength

(L17) that you do not know, add (them) together, (and) you see 2;20.

(L18) Halve 2;20, (and) you see 1;10. Square 1;10, (and) you see 1;21,40.

(L19) Add 1;21,40 to 0;34,4,26,40, (and)

(L20) you see 1;55,44,26,40.

(L21) What is the square root? 1;23,20 is the square root.square root

(L22) Subtract 1;10 (used in) completing the square from 1;23,20, and

(L23-24) you see 0;13,20. Make the reciprocal of 0;26,40, (and) you see 2;15.reciprocal of a number

(L25) Multiply 2;15 by 0;13,20, and

(L26) you see 0;30. 0;30 is the length.length
diagonalcompleting the square method

§.§.§ Mathematical Calculations

If we denote the lengthlength of four equal  quadrants of the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) by $x$, then  line  13 and equalities  <ref> and <ref>  provide us with  the following quadratic equationquadratic equation:

\begin{equation*}
(0;26,40) x^2 + x+(1;20)x = 1;16,40
\end{equation*}

\begin{equation}\label{equ-SMT20-g}
(0;26,40) x^2 +  (2;20)x = 1;16,40.
\end{equation}

Again,   according to lines 14-17, choose the new variable $y$ as  in <ref>
and multiply both sides of <ref> by $ 0;26,40$ to get a new quadratic equationquadratic equation with respect to $y$:

\begin{equation*}
\Big((0;26,40)\times x\Big)^2 + (2;20)\times\Big((0;26,40)\times x\Big) =(0;26,40)\times (1;16,40)
\end{equation*}

or equivalently

\begin{equation}\label{equ-SMT20-h}
y^2 + (2;20)y = 0;34,4,26,40.
\end{equation}

Next,  according to lines 18-23, we write

\begin{align*}
&~~  y^2 + (2;20)y = 0;34,4,26,40 \\
\Longrightarrow~~&~~    y^2 + 2\times\Big((1;10)\times y\Big)=  0;34,4,26,40 \\
\Longrightarrow~~&~~    y^2 + 2\times\Big((1;10)\times y\Big) +(1;10)^2=  0;34,4,26,40+(1;10)^2 \\
\Longrightarrow~~&~~ (y+1;10)^2  =  0;34,4,26,40+1;21,40 \\
\Longrightarrow~~&~~ (y+1;10)^2  =  1;55,44,26,40 \\
\Longrightarrow~~&~~ y+1;10=   \sqrt{1;55,44,26,40}\\
\Longrightarrow~~&~~ y+1;10=   \sqrt{(1;23,20)^2}\\
\Longrightarrow~~&~~  y+1;10= 1;23,20\\
\Longrightarrow~~&~~  y= 1;23,20-1;10\\
\Longrightarrow~~&~~   y= 0;13,20
\end{align*}

giving the solution of equation <ref>.
Finally,   according to lines 23-26  and similar to the first problem,  <ref>  implies that

\[x= \dfrac{1}{(0;26,40)}  \times (0;13,20)=(2;15)\times (0;13,20)=0;30\]

which is the solution of equation <ref>.

§.§ SMT No. 21: First Problem

§.§.§ Transliteration

Obverse:  Lines 1-21

(L1)  [a-pu-sà-mi-kum] 5-ta-àm uş$_4$-şa-am-m[a nigin]

(L2) [ab-ni] hi-[in-q]u sag i-na li-ib-bi il-[la-wi]

(L3) [35] a-šà dal-ba-na nigin-[i]a mi-nu {nu}

(L4) [za-e] 5 mi-i[s]-si$_{20}$-ta a-na 2 nigin 10 ta-mar

(L5) [10] nigin 1,40 ta-mar 1,40 i-na 35 zi 32(sic),20 ta-mar

(L6) 1 a-pu-sà-mi-ka gar 1,20 dal šà a-pu-sà-mi-ki gar

(L7) 1,20 a-na 1 i-ší-ma 1,20 ta-mar 1,20 ki-ma a-šà gar

(L8) tu-úr 1,20 nigin 1,46,40 ta-mar 1 nigin 1 ta-mar

(L9) 1 a-na 26,40 igi-gub a-pu-sà-mi-ki i-ší-ma

(L10) 26,40 ta-mar 26,40 i-na 1,46,40 zi

(L11) 1,20 ta-mar 1,20 a-na 33,20 a-šà dal-ba-na i-ší-ma

(L12) [4]4,26,40 ta-mar 1,[20] a-na 10 mi-is-si$_{20}$-ti i-ší-ma

(L13) [13,20] ta-mar 13,20 nigin [2,5]7,46,40 ta-mar

(L14) [2,57,4]6,40 a-na 4[4],26,40 dah

(L15) [47,24,2]6,40 t[a-mar] mi-na íb-si

(L16) [53,20 íb-si 13,20] ta-ki-il-ta-ka

(L17) [i-na 53,20 zi 40 ta-mar igi-1],20 šà ki-ma a-šà gar

(L18) [pu-ţú-úr 45 ta-mar 45 a-na 40] i-ší-ma

(L19) [30 ta-mar uš a-pu-sà-mi-ki 30 a-na 1,20 dal]

(L20) [šà a-pu-sà-mi-ki i-ší-ma 40 ta-mar 40 dal]

(L21) [10 mi-is-si$_{20}$-ta a-na 40 dah 50 ta-mar 50 nigin]

§.§.§ Translation

Obverse:  Lines 1-21

(L1) The (figure called) qplapusamikkum. I went (text: go) out by 5 (nindannindan (length unit)), and I made a square (field).lengthapusamikkum@apusamikkum (geometrical figure)

(L2) The narrow (that is, the qplapusamikkum) is enclosed therein with the sides (of the inner square).apusamikkum@apusamikkum (geometrical figure)

(L3) 35,0 is the area of the space between. What is my side of the (outer) square?area of an qplapusamikkum

(L4) You, multiply 5, the distance (between the outer and inner squares), by 2, (and) you see 10.

(L5) Square 10, (and) you see 1,40.  Subtract 1,40 from 35,0, (and) you see 33,20.

(L6) Put down 1 for the qplapusamikkum. Put down 1;20, the diagonal of qplapusamikkum.apusamikkum@apusamikkum (geometrical figure)

(L7) Multiply 1;20 by 1, and you see 1;20. Put down 1;20 like the area (of the inner square).area of a square

(L8) Return. Square 1;20, (and) you see 1;46,40. Square 1, (and) you see 1.

(L9) Multiply 1 by 0;26,40, the constant of qplapusamikkum, and

(L10) you see 0;26,40. Subtract 0;26,40 from 1;46,40, (and)

(L11) you see 1;20. Multiply 1;20 by 33,20, the (reduced) area of the space between, andarea of an apusamikkum

(L12) you see 44,26;40. Multiply 1;20 by 10 of the distance, and

(L13) you see 13;20. Square 13;20, (and) you see 2,57;46,40.

(L14) Add 2,57;46,40 to 44,26;40, (and)

(L15) you see 47,24;26,40. What is the square root?square root

(L16-17) 53;20 is the square root. Subtract 13;20, which was (used in) your completing the square, from 53;20, (and) you see 40. The reciprocal of 1;20,
{which you put down like the area,}reciprocal of a numbersquare rootarea of an apusamikkum

(L18) make (and) you see 0;45. Multiply 0;45 by 40, and

(L19-21) you see 30. (This is) the length of qplapusamikkum. Multiply 30 by 1;20, the diagonal of qplapusamikkum, and you see 40. 40 is the diagonal. Add 10, the distance, to 40, (and) you see 50. 50 is the side of the (outer) square.length
diagonalcompleting the square method

§.§.§ Mathematical Calculations

The   first problem involves the  figure qplapusamikkumapusamikkum@apusamikkum (geometrical figure)  already mentioned  in SMT No. 3 and SMT No. 20SMT No.t@SMT No. 20 (see [4] and <ref>). By looking carefully at the statement of the problem in lines 1-3, we realize that   the scribe first   considers an qplapusamikkumapusamikkum@apusamikkum (geometrical figure)   and then enlarges each side of the boundary squaresquare by 5 nindannindan (length unit) in each direction  to get a bigger squaresquare (see <ref>). Next, he assumes that the areaarea of a squarearea of an apusamikkum between the outer squaresquare and the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) to be 35,0. Finally, he asks for the value of the side of the bigger squaresquare.

Enlarging the boundary square of an qplapusamikkum  in four directions

As we will see shortly, all these data result in solving a specific  quadratic equationquadratic equation.     Recall that we can construct  an  qplapusamikkumapusamikkum@apusamikkum (geometrical figure) by removing four equal quadrants from a squaresquare whose side is  double the common radius of the   quadrantsquadrant and whose  vertexes are     coincide with the origins of the quadrantsquadrant.

To formulate the problem properly, as   shown in <ref>, we denote the  lengthlength of each quadrantquadrant in the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) $EFGH$ by $x$ and the lengthlength of the bigger (or outer) squaresquare $ABCD$ by $y$.

The side of boundary square of an qplapusamikkum extended in four directions

It follows from the properties of  <ref>  and the Babylonian approximation $\pi \approx  3$Babylonian approximate value for $\pi$ that

\[ 4x= 2\pi r \Longrightarrow r=\frac{4x}{2\pi}\approx \frac{4x}{6}=\frac{2x}{3} \]

\[ y=2(r+5)=2\times\frac{2x}{3}+10=(1;20)x+10 \]

so we get

\begin{equation}\label{equ-SMT21-a}
\begin{dcases}
\end{dcases}
\end{equation}

These relations easily give us the areas of the qplapusamikkumapusamikkum@apusamikkum (geometrical figure) $ EFGH$ and the outer squaresquare $ABCD $ as follows  (see [5]):

\begin{equation}\label{equ-SMT21-b}
\begin{dcases}
\end{dcases}
\end{equation}

Now, according to the lines 1-3, we must have

\[ S_{ABCD} - S_{EFGH}=35,0\]

which provides us with  the following quadratic equationquadratic equation

\begin{equation}\label{equ-SMT21-c}
\Big((1;20)x+10\Big)^2 - (0;26,40)x^2=35,0.
\end{equation}

Note that although we have obtained an equation  with respect to $x$, the scribe actually is looking for the value of the side of the outer squaresquare, that is,  $y= (1;20)x+10$. The solving process for equation <ref> now begins after line 3. According to lines 4-11, we can simplify the left-hand side of <ref> as follows:

\begin{align*}
&~~  \Big((1;20)x+10\Big)^2 - (0;26,40)x^2=35,0 \\
\Longrightarrow~~&~~    \Big((1;20)x \Big)^2+2\times 10 \times  (1;20)x  +10^2 -  (0;26,40)x^2=35,0 \\
\Longrightarrow~~&~~ (1;46,40)  x^2+2\times 10 \times  (1;20)x  +1,40 -  (0;26,40)x^2=35,0 \\
\Longrightarrow~~&~~ (1;46,40-0;26,40)  x^2+2\times 10 \times  (1;20)x  =35,0 -    1,40 \\
\Longrightarrow~~&~~  (1;20)  x^2+2\times 10 \times  (1;20)x   =33,20
\end{align*}

so we get a new quadratic equationquadratic equation

\begin{equation}\label{equ-SMT21-d}
(1;20)  x^2+2\times 10 \times  (1;20)x  =33,20.
\end{equation}

From now on, the scribe  utilizes  completing the squarecompleting the square method technique     in order to find the value of $x$ satisfying <ref>. According to lines 11-19, we can write

\begin{align*}
&~~   (1;20)  x^2+2\times 10 \times  (1;20)x  =33,20 \\
\Longrightarrow~~&~~   (1;20)\times   (1;20)  x^2 +2\times  10 \times (1;20) \times  (1;20)x   =(33,20)\times (1;20) \\
\Longrightarrow~~&~~      \Big((1;20) x\Big)^2+2\times  (13;20) \times  (1;20)x   =44,26;40  \\
\Longrightarrow~~&~~      \Big((1;20) x\Big)^2+2\times  (13;20) \times  (1;20)x  +(13;20)^2  =44,26;40+(13;20)^2  \\
\Longrightarrow~~&~~      \Big((1;20) x\Big)^2+2\times  (13;20) \times  (1;20)x  +(13;20)^2  =44,26;40+2,57;46,40  \\
\Longrightarrow~~&~~   \Big((1;20) x+13;20\Big)^2   =47,24;26,40 \\
\Longrightarrow~~&~~    (1;20) x+13;20   = \sqrt{47,24;26,40} \\
\Longrightarrow~~&~~    (1;20) x+13;20   = \sqrt{(53;20)^2} \\
\Longrightarrow~~&~~    (1;20) x+13;20   = 53;20\\
\Longrightarrow~~&~~    (1;20) x  = 53;20-13;20 \\
\Longrightarrow~~&~~    (1;20) x  = 40 \\
\Longrightarrow~~&~~    x  = \dfrac{1}{(1;20)} \times 40 \\
\Longrightarrow~~&~~    x  = \left(0;45\right)\times 40
\end{align*}

which gives us

\begin{equation}\label{equ-SMT21-e}
\end{equation}

As we said before, the scribe is interested in finding  the value of $y$. So, according to lines 19-21, we can use <ref> and <ref> to write

\[  y=(1;20)x+10 =(1;20)\times (30)+10=40+10=50.\]

Thus, the side of the outer squaresquare is

As is seen  in the discussion above, the scribe is extending a squaresquare by adding to each side a value of 5 nindannindan (length unit) in both directions. He also uses the areas of these squaressquare in his calculations which  require him to use and probably  expand  the algebraic expression

$$ \Big((1;20)x+10\Big)^2$$

or equivalently

$$\Big((1;20)x+2\times 5\Big)^2.$$

The number $10= 2\times 5$ in this expression   suggests that the  scribes of Susascribes of Susa  knew the algebraic  identityalgebraic identity

\begin{equation}\label{equ-SMT21-f}
(a+2b)^2 = a^2+4ab+4b^2.
\end{equation}

Because  the square of $10=2\times 5$ is   equal to  the total areas of four small squaressquare with sides of lengthlength 5  in the corners of <ref>.  In fact, a geometric explanation for verifying  the algebraic identityalgebraic identity <ref> can be given as follows.  As is clear from <ref>,  the left-hand side of <ref> is the areaarea of a square of the outer squaresquare  in the   figure  whose side is $a+2b$, that is $(a+2b)^2 $.

Extending  a square in four directions

On the other hand, the outer squaresquare is the union of the inner squaresquare with side $a$, four small squaressquare with sides $b$ and four rectanglerectangles with  lengths $a$ and  widths $b$. Since the areas of these parts are respectively

\[    a^2,\  4b^2,\ \  \text{and}\  \  4ab \]

the algebraic identityalgebraic identity <ref> is easily verified.

§.§ SMT No. 21: Second Problem

§.§.§ Transliteration

Reverse:  Lines 1-9

(L1)  [$ \cdots $ $ \cdots $ $ \cdots $] ta-mar

(L2) [$ \cdots $ $ \cdots $ 41,40] ta-mar mi-n[a í]b-si

(L3) [50 íb-si 10 ta-ki-il-t]a-ka i-na 50 zi

(L4) [40 ta-mar igi-1,20 pu]-ţú-úr 45 ta-mar

(L5) [45 a-na 40 i]-ší-ma 30 ta-mar uš a-pu-sà-mi-ki

(L6) [30 a-na 1,20] dal šà a-pu-sà-mi-ik-ki i-ší-ma

(L7) 40 ta-mar 40 dal 10 mi-is-si$_{20}$-ta šà uš

(L8) a-na 40 dah 50 ta-mar uš 5 mi-$<$is$>$-si$_{20}$-ta

(L9) šà sag a-na 40 dah 45 ta-mar sag

§.§.§ Translation

Reverse:  Lines 1-9

(L1) $ \cdots $ $ \cdots $ $ \cdots $ you see $ \cdots $.

(L2) $ \cdots $ $ \cdots $ $ \cdots $ you see 41,40. What is the square root?square root

(L3) 50 is the square root. Subtract 10, which was (used in) your completing the square, from 50, (and)square root

(L4) you see 40. Make the reciprocal of 1;20, (and) you see 0;45.reciprocal of a number

(L5) Multiply 0;45 by 40, and you see 30. (This is) the length of qplapusamikkum.length

(L6) Multiply 30 by 1;20, the diagonal of qplapusamikkum, andapusamikkum@apusamikkum (geometrical figure)

(L7) you see 40. 40 is the diagonal. 10, the distance of the length,length

(L8-9) add to 40, (and) you see 50. (This is) the length. 5, the distance of the width, add to 40, (and) you see 45. (This is) the width.lengthwidth
diagonalcompleting the square method

§.§.§ Mathematical Calculations

Although the beginning part of the text on the reverse is lost, we can construct the second problem and its solution by analyzing the remaining   text and comparing it with the first problem. After studying the remaining part of the second problem,  we   suggest that   the scribe  is again enlarging the boundary squaresquare of an qplapusamikkumapusamikkum@apusamikkum (geometrical figure) by adding 5 nindannindan (length unit) to the sides. Unlike the first problem, this time he appears to   extends two sides in both directions by 5 nindannindan (length unit)  and add 5 nindannindan (length unit) to the other two sides  in only one direction  which   produces a rectanglerectangle instead of a squaresquare (see <ref>).

Enlarging the boundary square of an qplapusamikkum  in three directions

According to <ref>, let us  denote the lengthlength of each quadrantquadrant  or the side of the  our qplapusamikkum  by $x$ and the sides of the outer rectanglerectangle  $ABCD$ by $y$ and $z$.
Similar to the first problem, here the scribe assumes the areaarea of a rectanglearea of an apusamikkum between the rectanglerectangle  $ABCD$ and the  qplapusamikkumapusamikkum@apusamikkum (geometrical figure)  $EFGH$ to be the known value 30,50  and then asks for the sides of the rectanglerectangle.  In other words,

\[ S_{ABCD}-S_{\Lambda}=30,50. \]

The side of   boundary square of an qplapusamikkum  extended in three directions

Similar to the first problem, we have $r=(1;20)x$, so  it follows from <ref> that

\begin{equation}\label{equ-SMT21-h}
\begin{dcases}
\end{dcases}
\end{equation}

so the areaarea of a rectanglearea of an apusamikkum of the outer rectanglerectangle is

$$ S_{ABCD}=yz= \Big((1;20)x+5\Big) \Big((1;20)x+10\Big).$$

Since the areaarea of a rectanglearea of an apusamikkum between the outer rectanglerectangle and the  qplapusamikkumapusamikkum@apusamikkum (geometrical figure)   is equal to  30,50 and $S_{EFGH}=(0;26,40)x^2$, we get the following quadratic equationquadratic equation:

\begin{equation}\label{equ-SMT21-i}
\Big((1;20)x+5\Big) \Big((1;20)x+10\Big)-(0;26,40)x^2=30,50.
\end{equation}

To solve equation  <ref>, we first need to simplify its left-hand side. So we   write

\begin{align*}
&~~   \Big((1;20)x+5\Big) \Big((1;20)x+10\Big)-(0;26,40)x^2=30,50 \\
\Longrightarrow~~&~~  \Big((1;20)x\Big)^2 +10\times  [(1;20)x] + 5\times  [(1;20)x] +5\times 10-(0;26,40)x^2=30,50   \\
\Longrightarrow~~&~~    \Big((1;20)^2 - 0;26,40\Big)    x^2 + 15\times   (1;20)x   +   50 =30,50     \\
\Longrightarrow~~&~~   (1;46,40 - 0;26,40) x^2 +15\times  (1;20)x   =30,50-50      \\
\Longrightarrow~~&~~    (1;20) x^2 +15\times   (1;20)x    =30,0
\end{align*}

so we obtain the new quadratic equationquadratic equation

\begin{equation}\label{equ-SMT21-j}
(1;20) x^2 +15\times (1;20)x   =30,0.
\end{equation}

Similar to the first problem, the scribe  has used the usual method of completing the squarecompleting the square method to solve this quadratic. To do so, he needs to multiply both sides of  <ref>  by 1;20. So we have

\begin{align*}
&~~  (1;20) x^2 +15\times   (1;20)x   =30,0 \\
\Longrightarrow~~&~~  (1;20)^2 x^2 +15\times (1;20) \times  (1;20)x   =(1;20)\times(30,0)   \\
\Longrightarrow~~&~~    \Big((1;20)  x\Big)^2 +2\times (7;30) \times (1;20)  \times  (1;20)x    =40,0    \\
\Longrightarrow~~&~~    \Big((1;20)  x\Big)^2 +2\times 10 \times (1;20)x   =40,0    \\
\Longrightarrow~~&~~    \Big((1;20)  x\Big)^2 +2\times 10 \times  (1;20)x  +10^2  =40,0+10^2    \\
\Longrightarrow~~&~~    \Big((1;20)  x\Big)^2 +2\times 10 \times  (1;20)x   +10^2  =40,0+1,40    \\
\Longrightarrow~~&~~    \Big((1;20)  x+10\Big)^2   =41,40
\end{align*}

hence we get

\begin{equation}\label{equ-SMT21-k}
\Big((1;20)  x+10\Big)^2   =41,40.
\end{equation}

(Note that the value 41,40 appears in line 2 on the reverse.) Now, according to lines 3-5, the scribe finds the solution of <ref> as follows:

\begin{align*}
&~~  \Big((1;20)  x+10\Big)^2   =41,40 \\
\Longrightarrow~~&~~  (1;20)  x+10   =\sqrt{41,40}   \\
\Longrightarrow~~&~~  (1;20)  x+10   =\sqrt{50^2}   \\
\Longrightarrow~~&~~      (1;20)  x+10   =50   \\
\Longrightarrow~~&~~   (1;20)  x  =50-10      \\
\Longrightarrow~~&~~    (1;20)  x  =40    \\
\Longrightarrow~~&~~       x  = \dfrac{1}{(1;20)} \times 40    \\
\Longrightarrow~~&~~       x  = (0;45)\times 40    \\
\Longrightarrow~~&~~       x  = 30.
\end{align*}

Therefore, we get

\begin{equation}\label{equ-SMT21-l}
\end{equation}

Next, according to lines 6-9, the scribe tries to find the values of $y$ and $z$. We can do this by using <ref> and <ref> as follows:

\[ y=(1;20)x+5=(1;20)\times 30+5=40+5=45 \]

\[ z=(1;20)x+10=(1;20)\times 30+10=40+10=50. \]

§ CONCLUSION

The texts of SMT No. 5 and SMT No. 6   show  that  the    Susa scribes  of these two tablet  had good skills in working with quadratic equations. As only the statements of problems have been provided in these tablets, one may suggest that they   prepared these texts  for educational goals.  The scribes     have taken great  care   in these texts to make sure that a greater number is never subtracted from a smaller one and  to ensure that the biggest number in a sum is always mentioned first. These two facts (also observed by  Friberg  and Al-Rawi   in [3]) imply that the concept of negative numbers was not known to the Susa scribes.  In addition, the forms of the equations in  Parts 6-12 of  SMT No. 5 suggest that the concept of number 0  was not known to them either. In fact,  instead of using the standard form $x^2+ax-b=0$, they have considered the alternative form $x^2+ax=b$.

Although SMT No. 20 and SMT No. 21      deal  with quadratic equations, their problems  derive  a complicated geometric figure and the areas of its different  parts. This suggests   that  the Susa scribes not only worked with complicated geometric shapes and knew how to calculate their areas, but that  they could also combine   geometric   and  algebraic skills to formulate and solve  complicated equations. This suggests   it is reasonable to say that the scribes of Susa were using methods that later formed the basics of   “geometric algebra”[This term refers to  the representation of algebraic concepts through geometric figures. In other words, the squares with sides  of length  $a$ can be thought of as geometric representations of $a^2$; rectangles with sides of lengths $a$ and $b$ can be interpreted as the products $ab$; and relationships among such objects can be interpreted as algebraic equations (see [6]).] in Greek mathematics.

[1]
E. M. Bruins and M. Rutten,    Textes Math'ematiques de Suse [TMS], Librairie Orientaliste Paul Geuthner,   Paris, 1961.

[2]
D. M. Burton, The History of Mathematics: An Introduction, 6th Edition, McGraw-Hill,   2006.

[3]
J. Friberg and F.  Al-Rawi, New Mathematical Cuneiform Texts, Sources and Studies in the History of Mathematics and Physical Sciences, Springer, 2016.

[4]
N. Heydari and K. Muroi, Bisection of Trapezoids in Elamite Mathematics, arXiv:2211.11139 [math.HO], November 2022, 21 pages: <https://arxiv.org/abs/2211.11139>.

[5]
N. Heydari and K. Muroi, Circular Figures in Elamite Mathematics, arXiv:2212.12423 [math.HO], December 2022, 28 pages:  <https://arxiv.org/abs/2212.12423>.

[6]
V.   Katz, A History of Mathematics: An Introduction, Addison-Wesley,   Third Edition,   2009.

[7]
K. Muroi, Mathematical Term Takīltum and Completing the Square in Babylonian Mathematics, Historia Scientiarum Vol. 12 (2003), pp. 254–263.