• February 8th 2010, 01:44 PM
jessicagoode
Hi all,

I need help/directions on how to solve this problem:

http://i45.tinypic.com/208absx.png

Thank you much.
• February 8th 2010, 01:58 PM
Quacky
Just to clarify, do you mean $\sqrt3x-2$ or $\sqrt{3x}-2$ or $\sqrt{3x-2}$?
• February 8th 2010, 02:01 PM
jessicagoode
Oh, sorry, the third one.

The square root of 3x-2
• February 8th 2010, 02:12 PM
Quacky
Ok. Here's my attempt:

$x-\sqrt{3x-2}=4$
$x-4=\sqrt{3x-2}$
$(x-4)^2=\sqrt{3x-2}^2$
$x^2-8x+16=3x-2$
$x^2-11x+18=0$
$(x-9)(x-2)=0$
$x=2$ or $x=9$

x=2 is a solution for the negative result of http://www.mathhelpforum.com/math-he...d71684ae-1.gif, because when I square root a number, the result could be positive or negative. $(-2)^2=2^2$ 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.
• February 8th 2010, 02:13 PM
Quacky
Sorry about the latex errors, I've corrected them now.