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} \]