## Algebra: A Combined Approach (4th Edition)

$x=\dfrac{1}{2}$ and $x=-4$
$2x^{2}+7x=4$ Take out common factor $2$ from the left side of the equation: $2\Big(x^{2}+\dfrac{7}{2}x\Big)=4$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $2\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular case, $b=\dfrac{7}{2}$ $2\Big[x^{2}+\dfrac{7}{2}x+\Big(\dfrac{7}{4}\Big)^{2}\Big]=4+2\Big(\dfrac{7}{4}\Big)^{2}$ $2\Big(x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}\Big)=4+\dfrac{49}{8}$ $2\Big(x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}\Big)=\dfrac{81}{8}$ Factor the expression inside the parentheses, which is a perfect square trinomial: $2\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{8}$ Take the $2$ to divide the right side of the equation: $\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{16}$ Take the square root of both sides of the equation: $\sqrt{\Big(x+\dfrac{7}{4}\Big)^{2}}=\pm\sqrt{\dfrac{81}{16}}$ $x+\dfrac{7}{4}=\pm\dfrac{9}{4}$ Solve for $x$: $x=\dfrac{-7\pm9}{4}$ The two solutions are: $x=\dfrac{-7+9}{4}=\dfrac{1}{2}$ $x=\dfrac{-7-9}{4}=-4$