## Intermediate Algebra (6th Edition)

$x = \{ 1, \frac{1}{4}\}$
$\sqrt (5x-1) -\sqrt x + 2 = 3$ To separate the radicals add $-2$ both sides. $\sqrt (5x-1) -\sqrt x +2 - 2 = 3 - 2$ $\sqrt (5x-1) -\sqrt x = 1$ Squaring on both sides. $(\sqrt (5x-1) -\sqrt x)^{2} = 1^{2}$ $(5x-1) + x - 2 \sqrt (5x-1)\sqrt x = 1$ $6x-1 - 2 \sqrt (5x-1)\sqrt x = 1$ $6x-1 -1 = 2 \sqrt (5x-1)\sqrt x$ $6x-2 = 2 \sqrt (5x-1)\sqrt x$ $2(3x-1) = 2 \sqrt (5x-1)\sqrt x$ Dividing both sides by $2$ $(3x-1) = \sqrt (5x-1)\sqrt x$ Squaring on both sides. $(3x-1)^{2} = (\sqrt (5x-1)\sqrt x )^{2}$ $9x^{2}-6x+1 = x(5x-1)$ $9x^{2}-6x+1 = 5x^{2}-x$ $9x^{2}-6x+1 - 5x^{2}+x = 0$ $4x^{2}-5x+1 = 0$ By factoring, $(4x-1)(x-1)=0$ $x = 1$ or $x= \frac{1}{4}$ $x = \{ 1, \frac{1}{4}\}$ Substituting $x$ values in the given equation, $x= 1$ satisfies the equation. So, $1$ is the only solution.