Chapter A.6, Problem 108AYU

To determine

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

Answer to Problem 108AYU

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

Explanation of Solution

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

$x^2+\frac{2}{3}x=\frac{1}{3}$

Since the coefficient of $x^2$ is 1, we can complete the square on the left side by adding ${\left(\frac{1}{2}\cdot \frac{2}{3}\right)}^2=\frac{1}{9}$. Of course, in an equation, whatever we add to the left side must also be added to the right side. Add $\frac{1}{9}$ to both sides.

Therefore $x^2+\frac{2}{3}x+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$.

${\left(x+\frac{1}{3}\right)}^2=\frac{4}{9}$

$x+\frac{1}{3}=\pm \sqrt{\frac{4}{9}}$

$x+\frac{1}{3}=\pm \frac{2}{3}$

$x+\frac{1}{3}=\frac{2}{3}$, $x+\frac{1}{3}=-\frac{2}{3}$

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