Solving Quadratic Equations: Completing the Square

Why complete the square?

A quadratic equation contains an $x^2$ term and can be written as

$ax^2+bx+c=0,\qquad a\ne0.$

Some quadratics factorise conveniently, but many do not. Completing the square replaces the $x^2$ and terms with a squared bracket. Once that square is isolated, the equation can be solved by taking square roots.

The method works because expanding a squared bracket gives

$(x+p)^2=x^2+2px+p^2.$

To match $x^2+bx$, choose $2p=b$, so $p=b/2$. However, expanding the bracket also introduces . Subtracting the same amount keeps the expression unchanged:

$x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2.$

This is the central completing-the-square identity.

First rearrange the equation

A quadratic may not initially have zero on one side. Move all terms to one side and collect like terms before completing the square.

For example,

$x^2+10x+4=-5$

becomes

$x^2+10x+9=0.$

The rearrangement matters: moving $-5$ to the left adds 5, so the constant becomes 9.

Solving when the coefficient of $x^2$ is 1

To complete the square in $x^2+10x+9$, halve 10 to obtain 5:

$x^2+10x+9=(x+5)^2-25+9=(x+5)^2-16.$

The equation is therefore

$(x+5)^2-16=0.$

Now isolate the square:

$(x+5)^2=16.$

There are two numbers whose square is 16, so

$x+5=\pm4.$

This produces two branches:

$x=-5+4=-1$

or

$x=-5-4=-9.$

Thus the solutions are $x=-1$ and $x=-9$. The $\pm$ is essential: using only the positive square root would lose one solution.

When the coefficient of $x^2$ is not 1

Consider

$3x^2-12x-15=0.$

Before completing the square, factor 3 from the $x^2$ and terms:

$3(x^2-4x)-15=0.$

Inside the bracket, halve $-4$ to get $-2$:

$x^2-4x=(x-2)^2-4.$

Substitute this equivalent expression back into the equation:

$3\big((x-2)^2-4\big)-15=0,$

so

$3(x-2)^2-27=0.$

Then solve:

$3(x-2)^2=27,$

$(x-2)^2=9,$

$x-2=\pm3.$

Therefore $x=5$ or $x=-1$.

Notice that the outside 3 multiplies the whole expression $(x-2)^2-4$. Dropping it from the $-4$ would change the equation.

A general completed-square form

The same reasoning gives

$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.$

You do not need to begin every question with this general identity. Often it is clearer to factor out $a$ from the first two terms and then halve the coefficient of $x$ inside the bracket. The identity does, however, show that the method works for any quadratic with $a\ne0$.

Exact solutions involving surds

Completing the square does not require the isolated square to equal a square number. For example,

$x^2+6x+2=0$

becomes

$(x+3)^2-7=0.$

Hence

$(x+3)^2=7$

and

$x=-3\pm\sqrt7.$

These are exact solutions. Convert them to decimals only if the question requests an approximation.

How many real solutions are possible?

After completing the square, an equation has the form

$(x+p)^2=k.$

• If $k>0$, there are two real solutions: $x=-p\pm\sqrt{k}$.

Finally, solutions can be checked by substituting them into the original equation. This also helps detect sign errors made while rearranging or completing the square.