Answer
$x=\frac{1}{2}$
Work Step by Step
$\sqrt (6x-2) = \sqrt (2x+3)- \sqrt (4x-1)$
Squaring on both sides.
$(\sqrt (6x-2) )^{2}= (\sqrt (2x+3)- \sqrt (4x-1))^{2}$
Using $(A-B)^{2}=A^{2}+B^{2}-2AB$
$(6x-2) = (2x+3) + (4x-1) -2 (\sqrt (2x+3)\sqrt (4x-1))$
$6x-2 = 6x+2 -2 (\sqrt (2x+3)\sqrt (4x-1))$
$2 (\sqrt (2x+3)\sqrt (4x-1)) = 6x+2 -6x+2$
$2 (\sqrt (2x+3)\sqrt (4x-1)) = 4$
$ \sqrt (2x+3)\sqrt (4x-1) = 2$
Again, squaring on both sides,
$ (\sqrt (2x+3)\sqrt (4x-1))^{2} = 2^{2}$
$(2x+3)(4x-1) = 4$
$8x^{2}+12x-2x-3-4=0$
$8x^{2}+10x-3-4=0$
$8x^{2}+10x-7=0$
By factoring,
$(4x+7)(2x-1)=0$
$4x+7=0$ or $2x-1=0$
$4x=-7$ or $2x=1$
$x=-\frac{7}{4}$ or $x = \frac{1}{2}$
By checking the proposed solutions in the given equation $x=-\frac{7}{4}$ is not a solution. It is an extraneous solution and $x = \frac{1}{2}$ is the only solution.
