# How do you solve sqrtx+3sqrtx=sqrt(17x-4)?

May 28, 2017

$x = 4$
$\sqrt{x} + 3 \sqrt{x} = \sqrt{17 x - 4} \mathmr{and} \sqrt{x} \left(3 + 1\right) = \sqrt{17 x - 4}$ or
$4 \sqrt{x} = \sqrt{17 x - 4}$, squaring both sides we get,
$16 x = 17 x - 4 \mathmr{and} x = 4$ [Ans]