Hi all,

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

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

Thank you much.

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- Feb 8th 2010, 01:44 PMjessicagoodeSolve the radical equation
Hi all,

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

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

Thank you much. - Feb 8th 2010, 01:58 PMQuacky
Just to clarify, do you mean $\displaystyle \sqrt3x-2$ or $\displaystyle \sqrt{3x}-2$ or $\displaystyle \sqrt{3x-2}$?

- Feb 8th 2010, 02:01 PMjessicagoode
Oh, sorry, the third one.

The square root of 3x-2 - Feb 8th 2010, 02:12 PMQuacky
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 http://www.mathhelpforum.com/math-he...d71684ae-1.gif, 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. - Feb 8th 2010, 02:13 PMQuacky
Sorry about the latex errors, I've corrected them now.