# Completing the square

If you've got a quadratic equation on the form of

$ax^{2}+c=0$

Then you can solve the equation by using the square root of

$x=\pm \sqrt{\frac{-c}{a}}$

Example

$3x^{2}-243=0$

$3x^{2}=243$

$x^{2}=\frac{243}{3}$

$x^{2}=81$

$x=\pm \sqrt{81}$

$x=9\: \: or\: \: x=-9$

This method can only be used if b = 0. If we instead have an equation on the form of

$x^{2}+bx=0$

we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as

$x^{2}+bx + d=\left ( x+d \right )^{2}=0$

This process is called completing the square and the constant d we're adding is

$d=\left (\frac{b}{2} \right )^{2}$

Example

$x^{2}+12x=0$

We begin by finding the constant d that can be used to complete the square.

$d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36$

$x^{2}+12x+d=0+d\Rightarrow$

$x^{2}+12x+36=0+36\Rightarrow$

$\begin{pmatrix}x+6 \end{pmatrix}^{2}=36$

$\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36}$

$\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix}$

The completing the square method could of course be used to solve quadratic equations on the form of

$ax^{2}+bx+c=0$

In this case you will add a constant d that satisfy the formula

$d=\left ( \frac{b}{2} \right )^{2}-c$

## Video lesson

Solve the equation by completing the squares

$x^{2} - 3x - 10=0$