Answer
$x=\displaystyle \frac{10}{9}$ or $x=2$
Work Step by Step
We are given:
$\sqrt{4x+1}-\sqrt{x-1}=2$
$\sqrt{4x+1}=2+\sqrt{x-1}$
We square both sides:
$(\sqrt{4x+1})^{2}=(2+\sqrt{x-1})^{2}$
$4x+1=4+(x-1)+4\sqrt{x-1}$
$4x+1=3+x+4\sqrt{x-1}$
$3x-2=4\sqrt{x-1}$
We square both sides again:
$(3x-2)^{2}=(4\sqrt{x-1})^{2}$
$9x^{2}-12x+4=16(x-1)$
And distribute:
$9x^{2}-12x+4=16x-16$
$9x^{2}-28x+20=0$
And factor:
$(9x-10)(x-2)=0$
Use the zero-factor property by equating each factor to zero:
$(9x-10)=0$ or $(x-2)=0$
$x=\displaystyle \frac{10}{9}$ or $x=2$