SOLUTION: can u please show me how to do this problem: the square root of (2x+5) - the square root of (x-2) = 3 I got the answer of 2, but i believe there is another possible answer,

Algebra ->  Square-cubic-other-roots -> SOLUTION: can u please show me how to do this problem: the square root of (2x+5) - the square root of (x-2) = 3 I got the answer of 2, but i believe there is another possible answer,      Log On


   

Question 10593: can u please show me how to do this problem:

the square root of (2x+5) - the square root of (x-2) = 3
I got the answer of 2, but i believe there is another possible answer, and i dont know how to get it. thanks
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B5%29+-+sqrt%28x-2%29+=+3

First step is to isolate one of the radicals. I recommend adding sqrt%28x-2%29 to each side of the equation.
sqrt%282x%2B5%29+=+3+%2B+sqrt%28x-2%29+

Next, square both sides:
%28sqrt%282x%2B5%29%29%5E2+=+%283+%2B+sqrt%28x-2%29%29%5E2+
%28sqrt%282x%2B5%29%29%5E2+=+%283+%2B+sqrt%28x-2%29%29%2A+%283+%2B+sqrt%28x-2%29%29+
2x%2B5+=+9%2B6+%2Asqrt%28x-2%29+%2B+x-2

Combine like terms on the right side:
2x%2B+5+=+7%2B+x+%2B+6%2Asqrt%28x-2%29+

Subtract 7 and x from each side:
2x+%2B+5+-+x+-+7+=+6+%2A+sqrt%28x-2%29
x+-2+=+6%2A+sqrt%28x-2%29+

Now, square both sides again:
%28x-2%29%5E2+=+36%28x-2%29+
x%5E2+-+4x+%2B+4+=+36x+-+72
x%5E2+-+4x+-+36x+%2B4+%2B+72+=+0
x%5E2+-+40x+%2B+76+=+0

This just happens to factor!! It always does in these types of problems!!
(x-2)(x-38) = 0
x=2 or x = 38

You still have to check to see if it really works in the original equation.

Let x = 2
sqrt%282x%2B5%29+-+sqrt%28x-2%29+=+3
sqrt%284%2B5%29+-+sqrt%280%29+=+3 It checks:

Let x = 38
sqrt%2876%2B5%29+-+sqrt%2836%29+=+3
sqrt%2881%29+-+sqrt%2836%29+=+3
9+-+6+=+3 This one checks also.

R^2 at SCC