# Completing the square

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In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form

${\displaystyle ax^{2}+bx+c\,\!}$

to the form

${\displaystyle a(\cdots \cdots )^{2}+{\mbox{constant}}.\,}$

In this context, "constant" means not depending on x. The expression inside the parenthesis is of the form (x + constant). Thus

${\displaystyle ax^{2}+bx+c\,\!}$ is converted to
${\displaystyle a(x+h)^{2}+k\,}$

for some values of h and k.

Completing the square is used in

In mathematics, completing the square is considered a basic algebraic operation, and is often applied without remark in any computation involving quadratic polynomials. Completing the square is also used to derive the quadratic formula.

## Overview

### Background

There is a simple formula in elementary algebra for computing the square of a binomial:

${\displaystyle (x+p)^{2}\,=\,x^{2}+2px+p^{2}.\,\!}$

For example:

{\displaystyle {\begin{alignedat}{2}(x+3)^{2}\,&=\,x^{2}+6x+9&&(p=3)\\[3pt](x-5)^{2}\,&=\,x^{2}-10x+25\qquad &&(p=-5).\end{alignedat}}}

In any perfect square, the number p is always half the coefficient of x, and the constant term is equal to p2.

### Basic example

Consider the following quadratic polynomial:

${\displaystyle x^{2}+10x+28.\,\!}$

This quadratic is not a perfect square, since 28 is not the square of 5:

${\displaystyle (x+5)^{2}\,=\,x^{2}+10x+25.\,\!}$

However, it is possible to write the original quadratic as the sum of this square and a constant:

${\displaystyle x^{2}+10x+28\,=\,(x+5)^{2}+3.}$

This is called completing the square.

### General description

Given any monic quadratic

${\displaystyle x^{2}+bx+c,\,\!}$

it is possible to form a square that has the same first two terms:

${\displaystyle \left(x+{\tfrac {1}{2}}b\right)^{2}\,=\,x^{2}+bx+{\tfrac {1}{4}}b^{2}.}$

This square differs from the original quadratic only in the value of the constant term. Therefore, we can write

${\displaystyle x^{2}+bx+c\,=\,\left(x+{\tfrac {1}{2}}b\right)^{2}+k,}$

where k is a constant. This operation is known as completing the square. For example:

{\displaystyle {\begin{alignedat}{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{alignedat}}}

### Non-monic case

Given a quadratic polynomial of the form

${\displaystyle ax^{2}+bx+c\,\!}$

it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial.

Example:

{\displaystyle {\begin{aligned}3x^{2}+12x+27&=3(x^{2}+4x+9)\\&{}=3\left((x+2)^{2}+5\right)\\&{}=3(x+2)^{2}+15\end{aligned}}}

This allows us to write any quadratic polynomial in the form

${\displaystyle a(x-h)^{2}+k.\,\!}$

### Formula

The result of completing the square may be written as a formula. For the general case:[1]

${\displaystyle ax^{2}+bx+c\;=\;a(x-h)^{2}+k,\quad {\text{where}}\quad h=-{\frac {b}{2a}}\quad {\text{and}}\quad k=c-ah^{2}=c-{\frac {b^{2}}{4a}}.}$

Specifically, when a=1:

${\displaystyle x^{2}+bx+c\;=\;(x-h)^{2}+k,\quad {\text{where}}\quad h=-{\frac {b}{2}}\quad {\text{and}}\quad k=c-{\frac {b^{2}}{4}}.}$

The matrix case looks very similar:

${\displaystyle x^{\mathrm {T} }Ax+x^{\mathrm {T} }b+c=(x-h)^{\mathrm {T} }A(x-h)+k\quad {\text{where}}\quad h=-{\frac {1}{2}}A^{-1}b\quad {\text{and}}\quad k=c-{\frac {1}{4}}b^{\mathrm {T} }A^{-1}b}$

where ${\displaystyle A}$ has to be symmetric.

If ${\displaystyle A}$ is not symmetric the formulae for ${\displaystyle h}$ and ${\displaystyle k}$ have to be generalized to:

${\displaystyle h=-(A+A^{\mathrm {T} })^{-1}b\quad {\text{and}}\quad k=c-h^{\mathrm {T} }Ah=c-b^{\mathrm {T} }(A+A^{\mathrm {T} })^{-1}A(A+A^{\mathrm {T} })^{-1}b}$.

## Relation to the graph

{{#invoke:Multiple image|render}} In analytic geometry, the graph of any quadratic function is a parabola in the xy-plane. Given a quadratic polynomial of the form

${\displaystyle (x-h)^{2}+k\quad {\text{or}}\quad a(x-h)^{2}+k}$

the numbers h and k may be interpreted as the Cartesian coordinates of the vertex of the parabola. That is, h is the x-coordinate of the axis of symmetry, and k is the minimum value (or maximum value, if a < 0) of the quadratic function.

One way to see this is to note that the graph of the function ƒ(x) = x2 is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function ƒ(x − h) = (x − h)2 is a parabola shifted to the right by h whose vertex is at (h, 0), as shown in the top figure. In contrast, the graph of the function ƒ(x) + kx2 + k is a parabola shifted upward by k whose vertex is at (0, k), as shown in the center figure. Combining both horizontal and vertical shifts yields ƒ(x − h) + k = (x − h)2 + k is a parabola shifted to the right by h and upward by k whose vertex is at (hk), as shown in the bottom figure.

## Solving quadratic equations

Completing the square may be used to solve any quadratic equation. For example:

${\displaystyle x^{2}+6x+5=0,\,\!}$

The first step is to complete the square:

${\displaystyle (x+3)^{2}-4=0.\,\!}$

Next we solve for the squared term:

${\displaystyle (x+3)^{2}=4.\,\!}$

Then either

${\displaystyle x+3=-2\quad {\text{or}}\quad x+3=2,}$

and therefore

${\displaystyle x=-5\quad {\text{or}}\quad x=-1.}$

This can be applied to any quadratic equation. When the x2 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 factoring the equation, which is only reliable if the roots are rational, completing the square will find the roots of a quadratic equation even when those roots are irrational or complex. For example, consider the equation

${\displaystyle x^{2}-10x+18=0.\,\!}$

Completing the square gives

${\displaystyle (x-5)^{2}-7=0,\,\!}$

so

${\displaystyle (x-5)^{2}=7.\,\!}$

Then either

${\displaystyle x-5=-{\sqrt {7}}\quad {\text{or}}\quad x-5={\sqrt {7}},\,}$

so

${\displaystyle x=5-{\sqrt {7}}\quad {\text{or}}\quad x=5+{\sqrt {7}}.\,}$

In terser language:

${\displaystyle x=5\pm {\sqrt {7}}.\,}$

Equations with complex roots can be handled in the same way. For example:

${\displaystyle {\begin{array}{c}x^{2}+4x+5\,=\,0\\[6pt](x+2)^{2}+1\,=\,0\\[6pt](x+2)^{2}\,=\,-1\\[6pt]x+2\,=\,\pm i\\[6pt]x\,=\,-2\pm i.\end{array}}}$

### Non-monic case

For an equation involving a non-monic quadratic, the first step to solving them is to divide through by the coefficient of x2. For example:

${\displaystyle {\begin{array}{c}2x^{2}+7x+6\,=\,0\\[6pt]x^{2}+{\tfrac {7}{2}}x+3\,=\,0\\[6pt]\left(x+{\tfrac {7}{4}}\right)^{2}-{\tfrac {1}{16}}\,=\,0\\[6pt]\left(x+{\tfrac {7}{4}}\right)^{2}\,=\,{\tfrac {1}{16}}\\[6pt]x+{\tfrac {7}{4}}={\tfrac {1}{4}}\quad {\text{or}}\quad x+{\tfrac {7}{4}}=-{\tfrac {1}{4}}\\[6pt]x=-{\tfrac {3}{2}}\quad {\text{or}}\quad x=-2.\end{array}}}$

## Other applications

### Integration

Completing the square may be used to evaluate any integral of the form

${\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}}$

using the basic integrals

${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C\quad {\text{and}}\quad \int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C.}$

For example, consider the integral

${\displaystyle \int {\frac {dx}{x^{2}+6x+13}}.}$

Completing the square in the denominator gives:

${\displaystyle \int {\frac {dx}{(x+3)^{2}+4}}\,=\,\int {\frac {dx}{(x+3)^{2}+2^{2}}}.}$

This can now be evaluated by using the substitution u = x + 3, which yields

${\displaystyle \int {\frac {dx}{(x+3)^{2}+4}}\,=\,{\frac {1}{2}}\arctan \left({\frac {x+3}{2}}\right)+C.}$

### Complex numbers

Consider the expression

${\displaystyle |z|^{2}-b^{*}z-bz^{*}+c,\,}$

where z and b are complex numbers, z* and b* are the complex conjugates of z and b, respectively, and c is a real number. Using the identity |u|2 = uu* we can rewrite this as

${\displaystyle |z-b|^{2}-|b|^{2}+c,\,\!}$

which is clearly a real quantity. This is because

{\displaystyle {\begin{aligned}|z-b|^{2}&{}=(z-b)(z-b)^{*}\\&{}=(z-b)(z^{*}-b^{*})\\&{}=zz^{*}-zb^{*}-bz^{*}+bb^{*}\\&{}=|z|^{2}-zb^{*}-bz^{*}+|b|^{2}.\end{aligned}}}

As another example, the expression

${\displaystyle ax^{2}+by^{2}+c,\,\!}$

where a, b, c, x, and y are real numbers, with a > 0 and b > 0, may be expressed in terms of the square of the absolute value of a complex number. Define

${\displaystyle z={\sqrt {a}}\,x+i{\sqrt {b}}\,y.}$

Then

{\displaystyle {\begin{aligned}|z|^{2}&{}=zz^{*}\\&{}=({\sqrt {a}}\,x+i{\sqrt {b}}\,y)({\sqrt {a}}\,x-i{\sqrt {b}}\,y)\\&{}=ax^{2}-i{\sqrt {ab}}\,xy+i{\sqrt {ba}}\,yx-i^{2}by^{2}\\&{}=ax^{2}+by^{2},\end{aligned}}}

so

${\displaystyle ax^{2}+by^{2}+c=|z|^{2}+c.\,\!}$

### Idempotent matrix

A matrix M is idempotent when M 2 = M. Idempotent matrices generalize the idempotent properties of 0 and 1. The completion of the square method of addressing the equation

${\displaystyle a^{2}+b^{2}=a,}$

shows that the idempotent 2 × 2 matrices are parametrized by a circle in the (a,b)-plane.

The matrix ${\displaystyle {\begin{pmatrix}a&b\\b&1-a\end{pmatrix}}}$ will be idempotent provided ${\displaystyle a^{2}+b^{2}=a,}$ which, upon completing the square, becomes

${\displaystyle (a-{\tfrac {1}{2}})^{2}+b^{2}={\tfrac {1}{4}}.}$

In the (a,b)-plane, this is the equation of a circle with center (1/2, 0) and radius 1/2.

## Geometric perspective

Consider completing the square for the equation

${\displaystyle x^{2}+bx=a.\,}$

Since x2 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 x2 and the bx rectangles into a larger square result in a missing corner. The term (b/2)2 added to each side of the above equation is precisely the area of the missing corner, whence derives the terminology "completing the square". [1]

## A variation on the technique

As conventionally taught, completing the square consists of adding the third term, v 2 to

${\displaystyle u^{2}+2uv\,}$

to get a square. There are also cases in which one can add the middle term, either 2uv or −2uv, to

${\displaystyle u^{2}+v^{2}\,}$

to get a square.

### Example: the sum of a positive number and its reciprocal

By writing

{\displaystyle {\begin{aligned}x+{1 \over x}&{}=\left(x-2+{1 \over x}\right)+2\\&{}=\left({\sqrt {x}}-{1 \over {\sqrt {x}}}\right)^{2}+2\end{aligned}}}

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

${\displaystyle x^{4}+324.\,\!}$

This is

${\displaystyle (x^{2})^{2}+(18)^{2},\,\!}$

so the middle term is 2(x2)(18) = 36x2. Thus we get

{\displaystyle {\begin{aligned}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{aligned}}}

(the last line being added merely to follow the convention of decreasing degrees of terms).

## References

1. {{#invoke:citation/CS1|citation |CitationClass=book }}, Section Formula for the Vertex of a Quadratic Function, page 133–134, figure 2.4.8
• 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