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Math problem  solving for a variable 
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#1
Aug1310, 07:22 AM

P: 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 4x3=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. Thank you for your time. 


#2
Aug1310, 07:30 AM

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Your mistake is in thinking (ab)^{2} = a^{2}b^{2}, which is what you did when you squared the lefthand side of the equation. You need to multiply it out correctly.



#3
Aug1310, 07:56 AM

P: 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
Aug1310, 08:08 AM

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Math problem  solving for a variable
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} = (ab)^{2} Now FOIL out the righthand side and then substitute back in for a and b. 


#5
Aug1310, 08:19 AM

P: 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
Aug1310, 08:31 AM

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I edited my previous post and took out the images. You should be able to read it now.



#7
Aug1310, 01:25 PM

P: 145

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


#8
Aug1310, 01:34 PM

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Looks fine.



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