Answer
$x=0$ or $x=3$
Work Step by Step
We are given:
$\sqrt{3x}=\sqrt{5x+1}-1$
We square both sides:
$(\sqrt{3x})^{2}=(\sqrt{5x+1}-1)^{2}$
$3x=5x+1-2\sqrt{5x+1}+1$
$3x=5x+2-2\sqrt{5x+1}$
$2\sqrt{5x+1}=2+5x-3x$
$2\sqrt{5x+1}=2+2x$
$\sqrt{5x+1}=1+x$
We square both sides again:
$(\sqrt{5x+1})^{2}=(1+x)^{2}$
$5x+1=1+2x+x^{2}$
$x^{2}-3x=0$
And factor:
$x(x-3)=0$
Use the zero-factor property by equating each factor to zero:
$x=0$ or $x-3=0$
$x=0$ or $x=3$
