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

$x=2$ and $x=-\dfrac{2}{3}$
$3x^{2}-4x=4$ Take out common factor $3$ from the left side of the equation: $3(x^{2}-\dfrac{4}{3}x)=4$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $3\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. In this particular problem, $b=-\dfrac{4}{3}$ $3\Big[x^{2}-\dfrac{4}{3}x+\Big(-\dfrac{2}{3}\Big)^{2}\Big]=4+3\Big(-\dfrac{2}{3}\Big)^{2}$ $3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=4+3\Big(\dfrac{4}{9}\Big)$ $3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=4+\dfrac{4}{3}$ $3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=\dfrac{16}{3}$ Factor the expression inside the parentheses, which is a perfect square trinomial: $3\Big(x-\dfrac{2}{3}\Big)^{2}=\dfrac{16}{3}$ Take the $3$ to divide the right side of the equation: $\Big(x-\dfrac{2}{3}\Big)^{2}=\dfrac{16}{9}$ Take the square root of both sides of the equation: $\sqrt{\Big(x-\dfrac{2}{3}\Big)^{2}}=\sqrt{\dfrac{16}{9}}$ $x-\dfrac{2}{3}=\pm\dfrac{4}{3}$ Solve for $x$: $x=\dfrac{2\pm4}{3}$ The two solutions are: $x=\dfrac{2+4}{3}=2$ $x=\dfrac{2-4}{3}=-\dfrac{2}{3}$