Hi all,
I need help/directions on how to solve this problem:
Thank you much.
Ok. Here's my attempt:
$\displaystyle x-\sqrt{3x-2}=4$
$\displaystyle x-4=\sqrt{3x-2}$
$\displaystyle (x-4)^2=\sqrt{3x-2}^2$
$\displaystyle x^2-8x+16=3x-2$
$\displaystyle x^2-11x+18=0$
$\displaystyle (x-9)(x-2)=0$
$\displaystyle x=2 $ or $\displaystyle x=9$
x=2 is a solution for the negative result of , because when I square root a number, the result could be positive or negative. $\displaystyle (-2)^2=2^2$ so square rooting also gives two possible solutions. The reason I bring this up is that otherwise, when checking it does not seem to work.