## Intermediate Algebra (12th Edition)

$\dfrac{1}{64}$
$\bf{\text{Solution Outline:}}$ To determine the number that will complete the square to solve the given equation, $4z^2-z-39=0 ,$ use first the properties of equality to express the equation in the form $x^2+bx=c.$ Once in this form, the needed number to complete the square of the left side is equal to $\left( \dfrac{b}{2} \right)^2.$ $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{4z^2-z-39}{4}=\dfrac{0}{4} \\\\ z^2-\dfrac{1}{4}z-\dfrac{39}{4}=0 \\\\ z^2-\dfrac{1}{4}z=\dfrac{39}{4} .\end{array} In the equation above, $b= -\dfrac{1}{4} .$ Using $\left( \dfrac{b}{2} \right)^2$, the number that will complete the square on the left side of the equal sign is \begin{array}{l}\require{cancel} \left( \dfrac{-\dfrac{1}{4}}{2} \right)^2 \\\\= \left( -\dfrac{1}{4}\div2 \right)^2 \\\\= \left( -\dfrac{1}{4}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( -\dfrac{1}{8} \right)^2 \\\\= \dfrac{1}{64} .\end{array}