## Intermediate Algebra (6th Edition)

$4$
$x-\sqrt (17-4x) -3 = 0$ $x-3 = \sqrt (17-4x)$ Squaring on both sides. $(x-3)^{2} = 17-4x$ Using $(a-b)^{2} = a^{2} -2ab +b^{2}$ $(x-3)^{2} = 17-4x$ $x^{2}-6x+9= 17-4x$ $x^{2}-6x+9- 17+4x=0$ $x^{2}-2x-8=0$ By factoring, $(x-4)(x+2)=0$ $x=4$ or $x=-2$ Substituting $x=4$ and $x=-2$ in the given equation, $x=4$ only satisfies the equation. $x=-2$ does not satisfy. So, $x=4$ is the solution.