Completing the square

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.

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.

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 ${\displaystyle k=c-{\frac {b^{2}}{4}}}$. This operation is known as completing the square. For example:

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

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:

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

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:

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

This allows the writing of any quadratic polynomial in the form

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

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

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

with

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

In particular, when a = 1, one has

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

with

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

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:

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

The matrix case looks very similar:

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

where ${\textstyle h=-{\frac {1}{2}}A^{-1}b}$ and ${\textstyle k=c-{\frac {1}{4}}b^{\mathrm {T} }A^{-1}b}$. Note that ${\displaystyle A}$ has to be symmetric.

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

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

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

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

so

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

Then either

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

In terser language:

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

so

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

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

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

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.

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

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

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

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

so

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

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

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

shows that some 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.

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.

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).

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

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

(Also known as Sophie Germain's identity).