Figure

EXAMPLE 3Completing the Square

Determine the number that must be added to each expression to complete the square. Then factor.

Start	Add	Result	Factored Form
$y^{2}+8y$	$\left( \dfrac{1}{2}\cdot 8\right) ^{\!\!2}=16$	$y^{2}+8y+16$	$( y+4) ^{2}$
$a^{2}-20a$	$\left( \dfrac{1}{2}\cdot ( -20) \right) ^{\!\!2}=100$	$a^{2}-20a+100$	$( a-10) ^{2}$
$p^{2}-5p$	$\left( \dfrac{1}{2}\cdot ( -5) \right) ^{\!\!2}= \dfrac{25}{4}$	$p^{2}-5p+\dfrac{25}{4}$	$\left( p-\dfrac{5}{2} \right) ^{\!\!2}$
$2x^{2}+6x=2 ( x^{2}+3x)$	$\left( \dfrac{1}{2}\cdot 3\right) ^{\!\!2}=\dfrac{9}{4}$	$2\!\left( x^{2}+3x+\dfrac{9}{4}\right)$	$2\!\left( x+\dfrac{3}{2}\right) ^{\!\!2}$

A-3

CAUTION The original expression $x^{2}+bx$ and the perfect square $x^{2}+bx+\left( \dfrac{b}{2}\right) ^{\!\!2}$ are not equal. So when completing the square within an equation or an inequality, we must not only add $\left( \dfrac{b}{2}\right) ^{\!\!2}\!$ , we must also subtract it. That is, $x^{\!\!2}+bx=x^{\!\!2}+bx+{\hbox{$\stackrel{\underbrace{\left(\dfrac{b}{2}\right) ^{\!\!2}-\left( \dfrac{b}{2}\right) ^{\!\!2}}}{=0}$}}=\left( x+\dfrac{b}{2}\right) ^{\!\!2}-\left( \dfrac{b}{2}\right) ^{\!\!2}$