Answer
$x_1=5$
$x_2=41$
Work Step by Step
Square each side:
$(\sqrt{2x-1}-\sqrt{x-5})^2=3^2$
$2x-1-2\sqrt{2x-1}\sqrt{x-5}+x-5=9$
Add same term variables and move the radicals to the right side:
$3x-6-2\sqrt{2x-1}\sqrt{x-5}-9+2\sqrt{2x-1}\sqrt{x-5}=9-9+2\sqrt{2x-1}\sqrt{x-5}$
$3x-15=2\sqrt{2x-1}\sqrt{x-5}$
Square each side again:
$(3x-15)^2=(2\sqrt{2x-1}\sqrt{x-5})^2$
$9x^2-45x-45x+225=4(2x-1)(x-5)$
Simplify:
$9x^2-90x+225=4(2x^2-10x-x+5)$
$9x^2-90x+225=4(2x^2-11x+5)$
$9x^2-90x+225=8x^2-44x+20$
$9x^2-90x+225-(8x^2-44x+20)=8x^2-44x+20-(8x^2-44x+20)$
$x^2-46x+205x=0$
Breaking down 205, we find that it only has two multiples, 41 and 5, which makes factoring possible:
$(x-5)(x-41)=0$
And we get two solutions:
$x_1-5=0$
$x_1=5$
$x_2-41=0$
$x_2=41$
We insert them into the original equation to find extraneous solutions:
$\sqrt{2(x_1)-1}-\sqrt{(x_1)-5}=3$
$\sqrt{2(5)-1}-\sqrt{(5)-5}=3$
$\sqrt{10-1}-\sqrt{0}=3$
$\sqrt{9}-0=3$
$3=3\checkmark$
$\sqrt{2(x_2)-1}-\sqrt{(x_2)-5}=3$
$\sqrt{2(41)-1}-\sqrt{(41)-5}=3$
$\sqrt{82-1}-\sqrt{36}=3$
$\sqrt{81}-6=3$
$9-6=3$
$3=3\checkmark$
