Chapter A.6, Problem 107AYU

To determine

To solve: Each equation by completing the square: $x^2-\frac{1}{2}x-\frac{3}{16}=0$.

Answer to Problem 107AYU

$x=\frac{3}{4},-\frac{1}{4}$

Explanation of Solution

Given, $x^2-\frac{1}{2}x-\frac{3}{16}=0$.

$x^2-\frac{1}{2}x=\frac{3}{16}$

Since the coefficient of $x^2$ is 1, we can complete the square on the left side by adding ${\left(\frac{1}{2}\cdot \frac{1}{2}\right)}^2=\frac{1}{16}$. Of course, in an equation, whatever we add to the left side must also be added to the right side. Add $\frac{1}{16}$ to both sides.

Therefore $x^2-\frac{1}{2}x+\frac{1}{16}=\frac{3}{16}+\frac{1}{16}$.

${\left(x-\frac{1}{4}\right)}^2=\frac{4}{16}$

${\left(x-\frac{1}{4}\right)}^2=\frac{1}{4}$

$x-\frac{1}{4}=\pm \sqrt{\frac{1}{4}}$

$x-\frac{1}{4}=\pm \frac{1}{2}$

$x-\frac{1}{4}=\frac{1}{2}$, $x-\frac{1}{4}=-\frac{1}{2}$

$x=\frac{1}{4}+\frac{1}{2}$, $x=\frac{1}{4}-\frac{1}{2}$

$x=\frac{3}{4}$, $x=-\frac{1}{4}$