SOLUTION: Please help solve this equation {{{ sqrt(2x+5)-sqrt(x-2)=3}}} I know you should isolate the radical so you add sqrt(x-2) to 3 it becomes: {{{Sqrt(2x+5)=3+sqrt(x-2)}}} and you th

Algebra ->  Radicals -> SOLUTION: Please help solve this equation {{{ sqrt(2x+5)-sqrt(x-2)=3}}} I know you should isolate the radical so you add sqrt(x-2) to 3 it becomes: {{{Sqrt(2x+5)=3+sqrt(x-2)}}} and you th      Log On


   

Question 964193: Please help solve this equation +sqrt%282x%2B5%29-sqrt%28x-2%29=3
I know you should isolate the radical so you add sqrt(x-2) to 3 it becomes:
Sqrt%282x%2B5%29=3%2Bsqrt%28x-2%29 and you then get rid of the radical so you have to square both side and then it cancles the radical. Now it's 2x+5=3+sqrt(x-2)^2 and then it's: 2x+5=3+sqrt(x-2)*3+sqrt(x-2) which equals: 2x+5=9+3sqrt(x-2)+3sqrt(x-2)+sqrt(x-2)^2 which cancels out and is just x-2.now we just combine like terms: 2x+5=7+6sqrt(x-2)+x now subtract 7 and x to 2x+5 and you get: x-2=6sqrt(x-2) and that's where I'm stuck I know you have to divide by 6 for both sides to isolate the radical and then square again but then what?I know the answer is x=2 and x=38 but how do you get it?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Sqrt%282x%2B5%29=3%2Bsqrt%28x-2%29

Square BOTH sides.
2x%2B5=%283%2Bsqrt%28x-2%29%29%5E2
2x%2B5=9%2B6sqrt%28x-2%29%2B%28x-2%29
Isolate the resulting radical.
2x%2B5-9-%28x-2%29=6sqrt%28x-2%29
2x-4-x%2B2=6sqrt%28x-2%29
x-2=6sqrt%28x-2%29
Square both sides again.
x%5E2-4x%2B4=36%28x-2%29
x%5E2-4x%2B4=36x-72
x%5E2-4x-36x%2B4%2B72=0
x%5E2-40x%2B76=0, which does not seem factorable.

x=%2840%2B-+sqrt%2840%5E2-4%2A76%29%29%2F2
x=%2840%2B-+sqrt%281904%29%29%2F2

1904=16*119

x=%2840%2B-+4sqrt%28119%29%29%2F2

highlight%28x=20%2B-+2sqrt%28119%29%29

(Possible mistake in this work)