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

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

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

In other words, completing the square places a perfect square trinomial inside of a quadratic expression.

Completing the square is used in

• solving quadratic equations,
• deriving the quadratic formula,
• graphing quadratic functions,
• evaluating integrals in calculus, such as Gaussian integrals with a linear term in the exponent,^[1]
• finding Laplace transforms.^[2][3]

In mathematics, completing the square is often applied in any computation involving quadratic polynomials.

History

The technique of completing the square was known in the Old Babylonian Empire.^[4]

Muhammad ibn Musa Al-Khwarizmi, a famous polymath who wrote the early algebraic treatise Al-Jabr, used the technique of completing the square to solve quadratic equations.^[5]

Overview

Background

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

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

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 coefficient of x is twice the number p, 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

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

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

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

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

${\displaystyle k=c-{\frac {b^{2)){4))}$

$${\displaystyle {\begin{alignedat}{1}x^{2}+6x+11\,&=\,(x+3)^{2}+2\\[3pt]x^{2}+14x+30\,&=\,(x+7)^{2}-19\\[3pt]x^{2}-2x+7\,&=\,(x-1)^{2}+6.\end{alignedat))}$$

Non-monic case

Given a quadratic polynomial of the form

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

Example:

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

Example:

$${\displaystyle {\begin{aligned}3x^{2}+12x+27&=3\left[x^{2}+4x\right]+27\\[1ex]&{}=3\left[(x+2)^{2}-4\right]+27\\[1ex]&{}=3(x+2)^{2}+3(-4)+27\\[1ex]&{}=3(x+2)^{2}-12+27\\[1ex]&{}=3(x+2)^{2}+15\end{aligned))}$$

This allows the writing of any quadratic polynomial in the form

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

Formula

Scalar case

The result of completing the square may be written as a formula. In the general case, one has^[6]

$${\displaystyle ax^{2}+bx+c=a(x-h)^{2}+k,}$$

$${\displaystyle h=-{\frac {b}{2a))\quad {\text{and))\quad k=c-ah^{2}=c-{\frac {b^{2)){4a)).}$$

In particular, when a = 1, one has

$${\displaystyle x^{2}+bx+c=(x-h)^{2}+k,}$$

$${\displaystyle h=-{\frac {b}{2))\quad {\text{and))\quad k=c-h^{2}=c-{\frac {b^{2)){4)).}$$

By solving the equation ${\displaystyle a(x-h)^{2}+k=0}$ in terms of ${\displaystyle x-h,}$ and reorganizing the resulting expression, one gets the quadratic formula for the roots of the quadratic equation:

$${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac))}{2a)).}$$

Matrix case

The matrix case looks very similar:

$${\displaystyle x^{\mathrm {T} }Ax+x^{\mathrm {T} }b+c=(x-h)^{\mathrm {T} }A(x-h)+k}$$

${\textstyle h=-{\frac {1}{2))A^{-1}b}$

${\textstyle k=c-{\frac {1}{4))b^{\mathrm {T} }A^{-1}b}$

${\displaystyle A}$

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

Graphs of quadratic functions shifted to the right by h = 0, 5, 10, and 15.

Graphs of quadratic functions shifted upward by k = 0, 5, 10, and 15.

Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.

In analytic geometry, the graph of any quadratic function is a parabola in the xy-plane. Given a quadratic polynomial of the form

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

One way to see this is to note that the graph of the function f(x) = x² is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function f(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 f(x) + k = x² + 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 f(x − h) + k = (x − h)² + k is a parabola shifted to the right by h and upward by k whose vertex is at (h, k), 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

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

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

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 reliable only 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

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

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

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

In terser language:

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

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

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

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

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:

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

Applying this procedure to the general form of a quadratic equation leads to the quadratic formula.

Other applications

Integration

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

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

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

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

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

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

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

As another example, the expression

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

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

Then

$${\displaystyle {\begin{aligned}|z|^{2}&{}=zz^{*}\\[1ex]&{}=\left({\sqrt {a))\,x+i{\sqrt {b))\,y\right)\left({\sqrt {a))\,x-i{\sqrt {b))\,y\right)\\[1ex]&{}=ax^{2}-i{\sqrt {ab))\,xy+i{\sqrt {ba))\,yx-i^{2}by^{2}\\[1ex]&{}=ax^{2}+by^{2},\end{aligned))}$$

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

Idempotent matrix

A matrix M is idempotent when M² = M. Idempotent matrices generalize the idempotent properties of 0 and 1. The completion of the square method of addressing the equation

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

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

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

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".^[7]

A variation on the technique

As conventionally taught, completing the square consists of adding the third term, v² to

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

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

Example: the sum of a positive number and its reciprocal

By writing

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

Example: factoring a simple quartic polynomial

Consider the problem of factoring the polynomial

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

This is

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

$${\displaystyle {\begin{aligned}x^{4}+324&{}=(x^{4}+36x^{2}+324)-36x^{2}\\&{}=(x^{2}+18)^{2}-(6x)^{2}={\text{a difference of two squares))\\&{}=(x^{2}+18+6x)(x^{2}+18-6x)\\&{}=(x^{2}+6x+18)(x^{2}-6x+18)\end{aligned))}$$

The same argument shows that ${\displaystyle x^{4}+4a^{4))$ is always factorizable as

$${\displaystyle x^{4}+4a^{4}=\left(x^{2}+2ax+2a^{2}\right)\left(x^{2}-2ax+2a^{2}\right)}$$

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

${\displaystyle ax^{3}+bx^{2}+cx+d}$

is a polynomial in x such that ${\displaystyle a\neq 0,}$ its two first terms are the two first terms of the expanded form of

${\displaystyle a\left(x+{\frac {b}{3a))\right)^{3}=ax^{3}+bx^{2}+x\,{\frac {b^{2)){3a))+{\frac {b^{3)){27a^{2))}.}$

So, the change of variable

${\displaystyle t=x+{\frac {b}{3a))}$

provides a cubic polynomial in ${\displaystyle t}$ without term of degree two, which is called the depressed form of the original polynomial.

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

More generally, a similar transformation can be used for removing terms of degree ${\displaystyle n-1}$ in polynomials of degree ${\displaystyle n.}$