Chapter A.6, Problem 109AYU

To determine

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

Answer to Problem 109AYU

$x=\frac{\sqrt{7}-1}{6},\frac{-1-\sqrt{7}}{6}$

Explanation of Solution

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

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

$x^2+\frac{1}{3}x=\frac{1}{6}$

Since the coefficient of $x^2$ is 1, we can complete the square on the left side by adding ${\left(\frac{1}{2}\cdot \frac{1}{3}\right)}^2=\frac{1}{36}$. Of course, in an equation, whatever we add to the left side must also be added to the right side. Add $\frac{1}{36}$ to both sides.

Therefore $x^2+\frac{1}{3}x+\frac{1}{36}=\frac{1}{6}+\frac{1}{36}$.

${\left(x+\frac{1}{6}\right)}^2=\frac{7}{36}$

$x+\frac{1}{6}=\pm \sqrt{\frac{7}{36}}$

$x+\frac{1}{6}=\pm \frac{\sqrt{7}}{6}$

$x+\frac{1}{6}=\frac{\sqrt{7}}{6}$, $x+\frac{1}{6}=-\frac{\sqrt{7}}{6}$

$x=\frac{\sqrt{7}-1}{6}$, $x=\frac{-1-\sqrt{7}}{6}$