## Precalculus: Mathematics for Calculus, 7th Edition

$x=\dfrac{1}{2}$ $x=-\dfrac{7}{2}$
$x^{2}+3x-\dfrac{7}{4}=0$ Let's take the independent term to the right side of the equation: $x^{2}+3x=\dfrac{7}{4}$ Remember that, in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation, $b=3$: $x^{2}+3x+(\dfrac{3}{2})^{2}=\dfrac{7}{4}+(\dfrac{3}{2})^{2}$ $x^{2}+3x+\dfrac{9}{4}=\dfrac{7}{4}+\dfrac{9}{4}$ $x^{2}+3x+\dfrac{9}{4}=\dfrac{16}{4}$ $x^{2}+3x+\dfrac{9}{4}=4$ We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes: $(x+\dfrac{3}{2})^{2}=4$ Take the square root of both terms: $\sqrt{(x+\dfrac{3}{2})^{2}}=\sqrt{4}$ $x+\dfrac{3}{2}=\pm2$ Solve for $x$: $x=-\dfrac{3}{2}\pm2$ The two solutions are: $x=-\dfrac{3}{2}+2=\dfrac{1}{2}$ and $x=-\dfrac{3}{2}-2=-\dfrac{7}{2}$