Editing Completing the square

{{short description|Method for solving quadratic equations}}
[[File:Completing the square.ogv|thumb|Animation depicting the process of completing the square. ([[:File:Completing the square.ogv|Details]], [[:File:Completing the square.gif|animated GIF version]])]]

In [[elementary algebra]], '''completing the square''' is a technique for converting a [[quadratic polynomial]] of the form {{tmath|\textstyle ax^2 + bx + c}} to the form {{tmath|\textstyle a(x-h)^2 + k}} for some values of {{tmath|h}} and {{tmath|k}}.<ref>{{cite book |title=Algebra: Themes, Tools, Concepts |author1=Anita Wah |author2=Creative Publications, Inc |edition= |publisher=Henri Picciotto |year=1994 |isbn=978-1-56107-251-4 |page=500 |url=https://books.google.com/books?id=nejW803k3fYC}} [https://books.google.com/books?id=nejW803k3fYC&pg=PA500 Extract of page 500] {{pb}}
{{cite book |title=Math Dictionary With Solutions |author1=Chris Kornegay |edition= |publisher=SAGE |year=1999 |isbn=978-0-7619-1785-4 |page=373 |url=https://books.google.com/books?id=Mjo5DQAAQBAJ}} [https://books.google.com/books?id=Mjo5DQAAQBAJ&pg=PA373 Extract of page 373] {{pb}}
The form <math>a(x+h)^2 + k</math> is also sometimes used. {{pb}} {{cite book |title=Cambridge IGCSE® Mathematics Core and Extended Coursebook |author1=Karen Morrison |author2=Nick Hamshaw |edition=illustrated, revised |publisher=Cambridge University Press |year=2018 |isbn=978-1-108-43718-9 |page=322 |url=https://books.google.com/books?id=6M1MDwAAQBAJ}} [https://books.google.com/books?id=6M1MDwAAQBAJ&pg=PA322 Extract of page 322] {{pb}} {{cite book |title=Foundation Mathematics for Engineers and Scientists with Worked Examples |author1=Shefiu Zakariyah |edition= |publisher=Taylor & Francis |year=2024 |isbn=978-1-003-85984-0 |page=254 |url=https://books.google.com/books?id=8IooEQAAQBAJ}} [https://books.google.com/books?id=8IooEQAAQBAJ&pg=PA254 Extract of page 254]</ref> In terms of a new quantity {{tmath|x-h}}, this expression is a quadratic polynomial with no linear term. By subsequently isolating {{tmath|\textstyle (x-h)^2}} and taking the [[square root]], a quadratic problem can be reduced to a linear problem.

The name ''completing the square'' comes from a geometrical picture in which {{tmath|x}} represents an unknown length. Then the quantity {{tmath|\textstyle x^2}} represents the area of a [[square]] of side {{tmath|x}} and the quantity {{tmath|\tfrac{b}{a}x}} represents the area of a pair of [[Congruence (geometry)|congruent]] [[rectangle]]s with sides {{tmath|x}} and {{tmath|\tfrac{b}{2a} }}. To this square and pair of rectangles one more square is added, of side length {{tmath|\tfrac{b}{2a} }}. This crucial step ''completes'' a larger square of side length {{tmath|x + \tfrac{b}{2a} }}.

Completing the square is the oldest method of solving general [[quadratic equation]]s, used in [[Old Babylonian Empire|Old Babylonian]] clay tablets dating from 1800–1600 BCE, and is still taught in elementary algebra courses today. It is also used for graphing [[quadratic function]]s, deriving the [[quadratic formula]], and more generally in computations involving quadratic polynomials, for example in [[calculus]] evaluating [[Gaussian integral]]s with a linear term in the exponent,<ref>{{cite book |title=Random Fields for Spatial Data Modeling: A Primer for Scientists and Engineers |author1=Dionissios T. Hristopulos |edition= |publisher=Springer Nature |year=2020 |isbn=978-94-024-1918-4 |page=267 |url=https://books.google.com/books?id=wivRDwAAQBAJ}} [https://books.google.com/books?id=wivRDwAAQBAJ&pg=PA267 Extract of page 267]</ref> and finding [[Laplace transform]]s.<ref>{{cite book |title=Differential Equations: An Introduction to Modern Methods and Applications |author1=James R. Brannan |author2=William E. Boyce |edition=3rd |publisher=John Wiley & Sons |year=2015 |isbn=978-1-118-98122-1 |page=314 |url=https://books.google.com/books?id=Sy2oDwAAQBAJ}} [https://books.google.com/books?id=Sy2oDwAAQBAJ&pg=PA314 Extract of page 314]</ref><ref>{{cite book |title=Introduction to Differential Equations with Dynamical Systems |author1=Stephen L. Campbell |author2=Richard Haberman |edition=illustrated |publisher=Princeton University Press |year=2011 |isbn=978-1-4008-4132-5 |page=214 |url=https://books.google.com/books?id=Mt3nI-lQKZQC}} [https://books.google.com/books?id=Mt3nI-lQKZQC&pg=PA214 Extract of page 214]</ref>

== History ==
{{Further|Algebra#History}}
The technique of completing the square was known in the [[Old Babylonian Empire]].<ref>Tony Philips, "[https://mathvoices.ams.org/featurecolumn/2020/11/01/fc-2020-10-2 Completing the Square]", ''American Mathematical Society Feature Column'', 2020.</ref>

[[Muhammad ibn Musa Al-Khwarizmi]], a famous [[polymath]] who wrote the early [[Algebra|algebraic]] treatise [[Al-jabr|Al-Jabr]], used the technique of completing the square to solve quadratic equations.<ref>{{Cite web |last=Hughes |first=Barnabas |date= |title=Completing the Square - Quadratics Using Addition |url=https://www.maa.org/press/periodicals/convergence/completing-the-square-quadratics-using-addition |access-date=2022-10-21 |website=[[Math Association of America]]}}</ref>

==Overview==

===Background===
The formula in [[elementary algebra]] for computing the [[square (algebra)|square]] of a [[binomial (polynomial)|binomial]] is:
<math display="block">(x + p)^2 \,=\, x^2 + 2px + p^2.</math>

For example:
<math display="block">\begin{alignat}{2}
(x+3)^2 \,&=\, x^2 + 6x + 9 && (p=3)\\[3pt]
(x-5)^2 \,&=\, x^2 - 10x + 25\qquad && (p=-5).
\end{alignat}
</math>

In any perfect square, the [[coefficient]] of ''x'' is twice the number ''p'', and the [[constant term]] is equal to ''p''<sup>2</sup>.

===Basic example===
Consider the following quadratic [[polynomial]]:
<math display="block">x^2 + 10x + 28.</math>

This quadratic is not a perfect square, since 28 is not the square of 5:
<math display="block">(x+5)^2 \,=\, x^2 + 10x + 25.</math>

However, it is possible to write the original quadratic as the sum of this square and a constant:
<math display="block">x^2 + 10x + 28 \,=\, (x+5)^2 + 3.</math>

This is called '''completing the square'''.

===General description===
Given any [[Monic polynomial|monic]] quadratic
<math display="block">x^2 + bx + c,</math>
it is possible to form a square that has the same first two terms:
<math display="block">\left(x+\tfrac{1}{2} b\right)^2 \,=\, x^2 + bx + \tfrac{1}{4}b^2.</math>

This square differs from the original quadratic only in the value of the constant
term. Therefore, we can write
<math display="block">x^2 + bx + c \,=\, \left(x + \tfrac{1}{2}b\right)^2 + k,</math>
where <math>k = c - \frac{b^2}{4}</math>. This operation is known as '''completing the square'''.
For example:
<math display="block">\begin{alignat}{1}
x^2 + 6x + 11 \,&=\, (x+3)^2 + 2 \\[3pt]
x^2 + 14x + 30 \,&=\, (x+7)^2 - 19 \\[3pt]
x^2 - 2x + 7 \,&=\, (x-1)^2 + 6.
\end{alignat}
</math>

===Non-monic case===
Given a quadratic polynomial of the form
<math display="block">ax^2 + bx + c</math>
it is possible to factor out the coefficient ''a'', and then complete the square for the resulting [[monic polynomial]].

Example:
<math display="block">
\begin{align}
  3x^2 + 12x + 27 &= 3[x^2+4x+9]\\
          &{}= 3\left[(x+2)^2 + 5\right]\\
          &{}= 3(x+2)^2 + 3(5)\\
          &{}= 3(x+2)^2 + 15
\end{align}</math>
This process of factoring out the coefficient ''a'' can further be simplified by only factorising it out of the first 2 terms. The integer at the end of the polynomial does not have to be included.

Example:
<math display="block">
\begin{align}
  3x^2 + 12x + 27 &= 3\left[x^2+4x\right] + 27\\[1ex]
          &{}= 3\left[(x+2)^2 -4\right] + 27\\[1ex]
          &{}= 3(x+2)^2 + 3(-4) + 27\\[1ex]
          &{}= 3(x+2)^2 - 12 + 27\\[1ex]
          &{}= 3(x+2)^2 + 15
\end{align}</math>

This allows the writing of any quadratic polynomial in the form
<math display="block">a(x-h)^2 + k.</math>

===Formula===
====Scalar case====
The result of completing the square may be written as a formula. In the general case, one has<ref>{{cite book
|last=Narasimhan
|first=Revathi
|year=2008
|title=Precalculus: Building Concepts and Connections
|publisher=Cengage Learning
|isbn=978-0-618-41301-0
|pages=133–134
|url=https://books.google.com/books?id=hLZz3xcP0SAC}}, [https://books.google.com/books?id=hLZz3xcP0SAC&pg=PA134 Section ''Formula for the Vertex of a Quadratic Function'', page 133–134,  figure 2.4.8]
</ref>
<math display="block">ax^2 + bx + c = a(x-h)^2 + k,</math>
with
<math display="block">h = -\frac{b}{2a} \quad\text{and}\quad k = c - ah^2 = c - \frac{b^2}{4a}.</math>

In particular, when {{math|1=''a'' = 1}}, one has
<math display="block">x^2 + bx + c = (x-h)^2 + k,</math>
with
<math display="block">h = -\frac{b}{2} \quad\text{and}\quad k = c - h^2 = c - \frac{b^2}{4}.</math>

By solving the equation <math>a(x-h)^2 + k=0</math> in terms of <math>x-h,</math> and reorganizing the resulting [[expression (mathematics)|expression]], one gets the [[quadratic formula]] for the roots of the [[quadratic equation]]:
<math display="block">x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.</math>

====Matrix case====
The [[matrix (mathematics)|matrix]] case looks very similar:
<math display="block">x^{\mathrm{T}}Ax + x^{\mathrm{T}}b + c = (x - h)^{\mathrm{T}}A(x - h) + k </math>
where <math display="inline"> h = -\frac{1}{2}A^{-1}b </math> and <math display="inline"> k = c - \frac{1}{4} b^{\mathrm{T}}A^{-1}b</math>. Note that <math>A</math> has to be [[symmetric matrix|symmetric]].

If <math>A</math> is not symmetric the formulae for <math>h</math> and <math>k</math> have to be generalized to:
<math display="block">h = -(A+A^{\mathrm{T}})^{-1}b \quad\text{and}\quad k = c - h^{\mathrm{T}}A h = c - b^{\mathrm{T}} (A+A^{\mathrm{T}})^{-1} A (A+A^{\mathrm{T}})^{-1}b</math>

==Relation to the graph==
{{multiple image
| align = right
| direction = vertical
| width = 200
| image1 = quartic h shift.svg
| alt1 = Graphs of quadratic functions shifted to the right by ''h'' = 0, 5, 10, and 15.
| caption1 = Graphs of quadratic functions shifted to the right by ''h'' = 0, 5, 10, and 15.
| image2 = quartic v shift.svg
| alt2 = Graphs of quadratic functions shifted upward by ''k'' = 0, 5, 10, and 15.
| caption2 = Graphs of quadratic functions shifted upward by ''k'' = 0, 5, 10, and 15.
| image3 = quartic hv shift.svg
| alt3 = Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.
| caption3 = Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.
}}
In [[analytic geometry]], the [[graph of a function|graph]] of any [[quadratic function]] is a [[parabola]] in the ''xy''-plane. Given a quadratic polynomial of the form
<math display="block">a(x-h)^2 + k</math>
the numbers ''h'' and ''k'' may be interpreted as the [[Cartesian coordinates]] of the [[Vertex (curve)|vertex]] (or [[stationary point]]) of the parabola. That is, ''h'' is the ''x''-coordinate of the axis of symmetry (i.e. the axis of symmetry has equation ''x = h''), and ''k'' is the [[maxima and minima|minimum value]] (or maximum value, if ''a''&nbsp;&lt;&nbsp;0) of the quadratic function.

One way to see this is to note that the graph of the [[function (mathematics)|function]] {{math|1=''f''(''x'') = ''x''<sup>2</sup>}} is a parabola whose vertex is at the origin&nbsp;(0,&nbsp;0). Therefore, the graph of the function {{math|1=''f''(''x'' &minus; ''h'') = (''x'' &minus; ''h'')<sup>2</sup>}} is a parabola shifted to the right by ''h'' whose vertex is at (''h'',&nbsp;0), as shown in the top figure. In contrast, the graph of the function {{math|1=''f''(''x'') + ''k'' = ''x''<sup>2</sup> + ''k''}} is a parabola shifted upward by {{mvar|k}} whose vertex is at {{math|(0, ''k'')}}, as shown in the center figure. Combining both horizontal and vertical shifts yields {{math|1=''f''(''x'' &minus; ''h'') + ''k'' = (''x'' &minus; ''h'')<sup>2</sup> + ''k''}} is a parabola shifted to the right by {{mvar|h}} and upward by {{mvar|k}} whose vertex is at {{math|(''h'', ''k'')}}, as shown in the bottom figure.

==Solving quadratic equations==
Completing the square may be used to solve any [[Quadratic equation#By completing the square|quadratic equation]]. For example:
<math display="block">x^2 + 6x + 5 = 0.</math>

The first step is to complete the square:
<math display="block">(x+3)^2 - 4 = 0.</math>

Next we solve for the squared term:
<math display="block">(x+3)^2 = 4.</math>

Then either
<math display="block">x+3 = -2 \quad\text{or}\quad x+3 = 2,</math>
and therefore
<math display="block">x = -5 \quad\text{or}\quad x = -1.</math>

This can be applied to any quadratic equation. When the ''x''<sup>2</sup> has a coefficient other than 1, the first step is to divide out the equation by this coefficient: for an example see the non-monic case below.

===Irrational and complex roots===
Unlike methods involving [[factorization|factoring]] the equation, which is reliable only if the roots are [[Rational number|rational]], completing the square will find the roots of a quadratic equation even when those roots are [[irrational number|irrational]] or [[Complex number|complex]]. For example, consider the equation
<math display="block">x^2 - 10x + 18 = 0.</math>

Completing the square gives
<math display="block">(x-5)^2 - 7 = 0,</math>
so
<math display="block">(x-5)^2 = 7.</math>
Then either
<math display="block">x-5 = -\sqrt{7} \quad\text{or}\quad x-5 = \sqrt{7}.</math>

In terser language:
<math display="block">x-5 = \pm \sqrt{7},</math>
so
<math display="block">x = 5 \pm \sqrt{7}.</math>

Equations with complex roots can be handled in the same way. For example:
<math display="block">\begin{align}
x^2 + 4x + 5 &= 0 \\[6pt]
(x+2)^2 + 1 &= 0 \\[6pt]
(x+2)^2 &= -1 \\[6pt]
x+2 &= \pm i \\[6pt]
x &= -2 \pm i.
\end{align}</math>

===Non-monic case===
For an equation involving a non-[[Monic polynomial|monic]] quadratic, the first step to solving them is to divide through by the coefficient of ''x''<sup>2</sup>. For example:

<math display="block">\begin{array}{c}
2x^2 + 7x + 6 \,=\, 0 \\[6pt]
x^2 + \tfrac{7}{2}x + 3 \,=\, 0 \\[6pt]
\left(x+\tfrac{7}{4}\right)^2 - \tfrac{1}{16} \,=\, 0 \\[6pt]
\left(x+\tfrac{7}{4}\right)^2 \,=\, \tfrac{1}{16} \\[6pt]
x+\tfrac{7}{4} = \tfrac{1}{4} \quad\text{or}\quad x+\tfrac{7}{4} = -\tfrac{1}{4} \\[6pt]
x = -\tfrac{3}{2} \quad\text{or}\quad x = -2.
\end{array}
</math>

Applying this procedure to the general form of a quadratic equation leads to the [[Quadratic formula#Derivations|quadratic formula]].

==Other applications==

===Integration===
Completing the square may be used to evaluate any integral of the form
<math display="block">\int \frac{dx}{ax^2+bx+c}</math>
using the basic integrals
<math display="block">\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| +C \quad\text{and}\quad
\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) +C.</math>

For example, consider the integral
<math display="block">\int \frac{dx}{x^2 + 6x + 13}.</math>

Completing the square in the denominator gives:
<math display="block">\int \frac{dx}{(x+3)^2 + 4} \,=\, \int\frac{dx}{(x+3)^2 + 2^2}.</math>

This can now be evaluated by using the [[integration by substitution|substitution]]
''u''&nbsp;=&nbsp;''x''&nbsp;+&nbsp;3, which yields
<math display="block">\int\frac{dx}{(x+3)^2 + 4} \,=\, \frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C.</math>

===Complex numbers===
Consider the expression
<math display="block"> |z|^2 - b^*z - bz^* + c,</math>
where ''z'' and ''b'' are [[complex number]]s, ''z''<sup>*</sup> and ''b''<sup>*</sup> are the [[complex conjugate]]s of ''z'' and ''b'', respectively, and ''c'' is a [[real number]]. Using the identity |''u''|<sup>2</sup> = ''uu''<sup>*</sup> we can rewrite this as
<math display="block"> |z-b|^2 - |b|^2 + c , </math>
which is clearly a real quantity. This is because
<math display="block">\begin{align}
  |z-b|^2 &{}=  (z-b)(z-b)^*\\
          &{}=  (z-b)(z^*-b^*)\\
          &{}= zz^* - zb^* - bz^* + bb^*\\
          &{}=  |z|^2 - zb^* - bz^* + |b|^2 .
\end{align}</math>

As another example, the expression
<math display="block">ax^2 + by^2 + c ,</math>
where ''a'', ''b'', ''c'', ''x'', and ''y'' are real numbers, with ''a''&nbsp;&gt;&nbsp;0 and ''b''&nbsp;&gt;&nbsp;0, may be expressed in terms of the square of the [[absolute value]] of a complex number. Define
<math display="block">z = \sqrt{a}\,x + i \sqrt{b} \,y .</math>

Then
<math display="block">\begin{align}
|z|^2
&{}= z z^*\\[1ex]
&{}= \left(\sqrt{a}\,x + i \sqrt{b}\,y\right) \left(\sqrt{a}\,x - i \sqrt{b}\,y\right) \\[1ex]
&{}= ax^2 - i\sqrt{ab}\,xy + i\sqrt{ba}\,yx - i^2 by^2 \\[1ex]
&{}= ax^2 + by^2 ,
\end{align}</math>
so
<math display="block"> ax^2 + by^2 + c = |z|^2 + c . </math>

===Idempotent matrix===
A [[matrix (mathematics)|matrix]] ''M'' is [[idempotent matrix|idempotent]] when ''M''<sup>2</sup> = ''M''. Idempotent matrices generalize the idempotent properties of 0 and 1. The completion of the square method of addressing the equation 
<math display="block">a^2 + b^2 = a ,</math>
shows that some idempotent 2×2 matrices are parametrized by a [[circle]] in the (''a'',''b'')-plane:

The matrix <math>\begin{pmatrix}a & b \\ b & 1-a \end{pmatrix}</math> will be idempotent provided <math>a^2 + b^2 = a ,</math> which, upon completing the square, becomes
<math display="block">(a - \tfrac{1}{2})^2 + b^2 = \tfrac{1}{4} .</math>
In the (''a'',''b'')-plane, this is the equation of a circle with center (1/2, 0) and radius 1/2.

==Geometric perspective==
[[Image:Completing the square.svg|right|250px]]
Consider completing the square for the equation
<math display="block">x^2 + bx = a.</math>

Since ''x''<sup>2</sup> represents the area of a square with side of length ''x'', and ''bx'' represents the area of a rectangle with sides ''b'' and ''x'', the process of completing the square can be viewed as visual manipulation of rectangles.

Simple attempts to combine the ''x''<sup>2</sup> and the ''bx'' rectangles into a larger square result in a missing corner. The term (''b''/2)<sup>2</sup> added to each side of the above equation is precisely the area of the missing corner, whence derives the terminology "completing the square".<ref>{{cite book | last1=Carroll | first1=Maureen T. | last2=Rykken | first2=Elyn | title=Geometry: The Line and the Circle | publisher=American Mathematical Society | series=AMS/MAA Textbooks | year=2018 | isbn=978-1-4704-4843-1 | url=https://books.google.com/books?id=CmuBDwAAQBAJ&pg=PA162 | access-date=2024-03-31 | page=162}}</ref>

==A variation on the technique==
As conventionally taught, completing the square consists of adding the third term, ''v''{{i sup|2}} to
<math display="block">u^2 + 2uv</math>
to get a square. There are also cases in which one can add the middle term, either 2''uv'' or &minus;2''uv'', to
<math display="block">u^2 + v^2</math>
to get a square.

===Example: the sum of a positive number and its reciprocal===
By writing
<math display="block">\begin{align}
x + {1 \over x} &{} = \left(x - 2 + {1 \over x}\right) + 2\\
                &{}= \left(\sqrt{x} - {1 \over \sqrt{x}}\right)^2 + 2
\end{align}</math>
we show that the sum of a positive number ''x'' and its reciprocal is always greater than or equal to 2. The square of a real expression is always greater than or equal to zero, which gives the stated bound; and here we achieve 2 just when ''x'' is 1, causing the square to vanish.

===Example: factoring a simple quartic polynomial===
Consider the problem of factoring the polynomial
<math display="block">x^4 + 324 . </math>

This is
<math display="block">(x^2)^2 + (18)^2, </math>
so the middle term is 2(''x''<sup>2</sup>)(18)&nbsp;=&nbsp;36''x''<sup>2</sup>. Thus we get
<math display="block">\begin{align}
x^4 + 324
&{}= (x^4 + 36x^2 + 324 ) - 36x^2  \\
&{}= (x^2 + 18)^2 - (6x)^2 =\text{a difference of two squares} \\
&{}= (x^2 + 18 + 6x)(x^2 + 18 - 6x) \\
&{}= (x^2 + 6x + 18)(x^2 - 6x + 18)
\end{align}</math>
(the last line being added merely to follow the convention of decreasing degrees of terms).

The same argument shows that <math>x^4 + 4a^4 </math> is always factorizable as 
<math display="block">x^4 + 4a^4 = \left(x^2+2a x + 2a^2\right) \left(x^2-2 ax + 2a^2\right)</math>
(Also known as [[Sophie Germain's identity]]).

==Completing the cube==

"Completing the square" consists to remark that the two first terms of a [[quadratic polynomial]] are also the first terms of the square of a [[linear polynomial]], and to use this for expressing the quadratic polynomial as the sum of a square and a constant. 

'''Completing the cube''' is a similar technique that allows to transform a [[cubic polynomial]] into a cubic polynomial without term  of degree two.

More precisely, if 
:<math>ax^3+bx^2+cx+d</math>
is a polynomial in {{mvar|x}} such that <math>a\ne 0,</math> its two first terms are the two first terms of the expanded form of 
:<math>a\left(x+\frac b {3a}\right)^3=ax^3+bx^2+x\,\frac{b^2}{3a}+\frac {b^3}{27a^2}.</math>

So, the [[change of variable]]
:<math>t=x+\frac b {3a}</math>
provides a cubic polynomial in <math>t</math> without term of [[Degree of a polynomial|degree]] two, which is called the [[depressed cubic|depressed form]] of the original polynomial.

This transformation is generally the first step of the methods for solving the general cubic equation.

More generally, a similar transformation can be used for removing terms of degree <math>n-1</math> in polynomials of degree <math>n</math>, which is called [[Tschirnhaus transformation]].

==References==
{{reflist}}
*Algebra 1, Glencoe, {{ISBN|0-07-825083-8}}, pages 539–544
*Algebra 2, Saxon, {{ISBN|0-939798-62-X}}, pages 214–214, 241–242, 256–257, 398–401

==External links==
{{commons category}}
*{{PlanetMath |urlname=completingthesquare |title=Completing the square}}


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