Posted by **Micha** on Monday, October 1, 2007 at 12:35am.

How do I work this correctly? I know the answer is 7 but I am doing something wrong.Here is what I have...

x+ã(2x-5)=10

x+ã(2x-5)^2=10^2

x+2x-5=100

3x-5+5=100+5

3x=105

x=35

PLEASE HELP!

- Algebra -
**Nick**, Monday, October 1, 2007 at 12:55am
what is the a~ all about? If it is a square root symbol then the solution looks like this:

x + (2x-5)^(1/2) = 10 : move x to other side

(2x-5)^(1/2) = 10 - x : square both sides

2x-5 = (10-x)^2 : expand the squared term

2x-5 = 100-10x-10x+x^2 : combine terms

0 = 105 - 22x + x^2 : use quadratic formula to solve for x

x =[-(-22) +/- sqrt(22^2 -4*1*105)]/(2*1)

x= [22 +/- sqrt(484 - 420)]/2

x= [22 +/- 8] / 2

x= 7 or 15

Plugging back into our initial equation shows that 7 is a solution, but 15 is not. This is because we started with a square root, which is invalid for negative numbers.

Regards.

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