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

$x=2$
$2x=\sqrt{10+3x}$ Square both sides of the equation: $(2x)^{2}=(\sqrt{10+3x})^{2}$ $4x^{2}=10+3x$ Take all terms to the left side: $4x^{2}-3x-10=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=4$, $b=-3$ and $c=-10$. Substitute: $x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(4)(-10)}}{2(4)}=\dfrac{3\pm\sqrt{9+160}}{8}=...$ $...=\dfrac{3\pm\sqrt{169}}{8}=\dfrac{3\pm13}{8}$ The two solutions are: $x=\dfrac{3+13}{8}=2$ $x=\dfrac{3-13}{8}=-\dfrac{5}{4}$ Finally, verify the solutions: $2(2)=\sqrt{10+3(2)}$ $4=4$ True $2\Big(-\dfrac{5}{4}\Big)=\sqrt{10+3\Big(-\dfrac{5}{4}\Big)}$ $-\dfrac{5}{2}=\dfrac{5}{2}$ False The answer is $x=2$