## Precalculus: Mathematics for Calculus, 7th Edition

$x=0$ and $x=\dfrac{1}{4}$
$4x^{2}-x=0$ First, take out common factor $4$: $4\Big(x^{2}-\dfrac{1}{4}x\Big)=0$ Take the $4$ to divide the right side of the equation. We get: $x^{2}-\dfrac{1}{4}x=0$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular problem, $b=-\dfrac{1}{4}$ $x^{2}-\dfrac{1}{4}x+\Big(\dfrac{-1/4}{2}\Big)^{2}=\Big(\dfrac{-1/4}{2}\Big)^{2}$ $x^{2}-\dfrac{1}{4}x+\dfrac{1}{64}=\dfrac{1}{64}$ Factor the perfect square trinomial that we have on the left side of the equation: $\Big(x-\dfrac{1}{8}\Big)^{2}=\dfrac{1}{64}$ Solve for $x$: $x-\dfrac{1}{8}=\pm\sqrt{\dfrac{1}{64}}$ $x=\dfrac{1}{8}\pm\dfrac{1}{8}$ So, our two solutions are: $x=\dfrac{1}{8}-\dfrac{1}{8}=0$ and $x=\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}$