# Math problem - solving for a variable

1. ### Matt1234

145
1. The problem statement, all variables and given/known data

sqrt(2x+4) = sqrt(6x+1) - 1

I Need to solve for x, but cannot seem to get the same answer as the text. (Ans. x= 5/2)

3. The attempt at a solution

sqrt(2x+4) - sqrt(6x+1) = -1
square both sides
(2x+4) - (6x+1) = 1
-4x-3=1
-4x=4
x=-1 ??

I know I must be doing something wrong, it has been a while since I have done this sort of problem.

2. ### vela

12,443
Staff Emeritus
Your mistake is in thinking (a-b)2 = a2-b2, which is what you did when you squared the lefthand side of the equation. You need to multiply it out correctly.

3. ### Matt1234

145
Im still not there yet:

My new attempt:

I cant seem to get rid of the sqrt (x)
i know sqrt(12x) = 2*sqrt(3x)

4. ### vela

12,443
Staff Emeritus
Now you're making even more mistakes. First,

[sqrt(2x+4)-sqrt(6x+1)]2 ≠ (2x+4) - (6x+1)

Second,

sqrt(a+b) ≠ sqrt(a)+sqrt(b)

which is what you're doing going from the fifth line to the sixth line. You were, however, correct when you said earlier that

(sqrt(2x+4))2 = 2x+4

The problem is actually a bit easier to solve if you square both sides right away:

(sqrt(2x+4))2 = (sqrt(6x+1)-1)2

To correctly calculate the righthand side, let a=sqrt(6x+1) and b=1. Then your equation becomes

(sqrt(2x+4))2 = (a-b)2

Now FOIL out the righthand side and then substitute back in for a and b.

Last edited: Aug 13, 2010
5. ### Matt1234

145
Im sorry, I am unable to see that. All i can see is black and faint white portions. Is it possible for you to take a screen shot and upload to image shack?

6. ### vela

12,443
Staff Emeritus
I edited my previous post and took out the images. You should be able to read it now.

7. ### Matt1234

145
I got it, thank you for your help.
It took a while to come back to me. I appreciate your time, thank you.

Looks fine.