SOLUTION: Solve for x: {{{sqrt(x+7) + sqrt(x+2) = sqrt(x-1) - sqrt(x-2)}}} When I solve the above equation, I get x = 2; but when I verify my answer by substituting the value of x in the

Algebra ->  Equations -> SOLUTION: Solve for x: {{{sqrt(x+7) + sqrt(x+2) = sqrt(x-1) - sqrt(x-2)}}} When I solve the above equation, I get x = 2; but when I verify my answer by substituting the value of x in the      Log On


   

Question 54627: Solve for x: sqrt%28x%2B7%29+%2B+sqrt%28x%2B2%29+=+sqrt%28x-1%29+-+sqrt%28x-2%29

When I solve the above equation, I get x = 2; but when I verify my answer by substituting the value of x in the original equation, it is not matching. I tried a few times to solve using different methods and I am always getting the same value of x as 2. What am I doing wrong?
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B7%29+%2B+sqrt%28x%2B2%29+=+sqrt%28x-1%29+-+sqrt%28x-2%29
Squaring both sides
%28sqrt%28x%2B7%29+%2B+sqrt%28x%2B2%29%29%5E2+=+%28sqrt%28x-1%29+-+sqrt%28x-2%29%29%5E2


12+%2B+2%2Asqrt%28x%2B7%29%2Asqrt%28x%2B2%29+=+-+2%2Asqrt%28x-1%29%2Asqrt%28x-2%29
6+=+-sqrt%28x%2B7%29%2Asqrt%28x%2B2%29+-+sqrt%28x-1%29%2Asqrt%28x-2%29
Squaring again



10+=+x%5E2%2B3x+%2B+sqrt%28x%2B7%29%2Asqrt%28x%2B2%29%2Asqrt%28x-1%29%2Asqrt%28x-2%29
x%5E2+-+3x+-+10=+-sqrt%28x%2B7%29%2Asqrt%28x%2B2%29%2Asqrt%28x-1%29%2Asqrt%28x-2%29
%28x%2B2%29%28x-5%29+=+-sqrt%28x%2B7%29%2Asqrt%28x%2B2%29%2Asqrt%28x-1%29%2Asqrt%28x-2%29
Squaring again
%28x%2B2%29%5E2%28x-5%29%5E2+=+%28x%2B7%29%28x%2B2%29%28x-1%29%28x-2%29
%28x%2B2%29%28%28x%2B2%29%28x-5%29%5E2+-+%28x%2B7%29%28x-1%29%28x-2%29%29+=+0
%28x%2B2%29%28%28x%2B2%29%28x%5E2+%2B+10x+%2B+25%29+-+%28x%5E2%2B6x-7%29%28x-2%29%29+=+0

%28x%2B2%29%288x%5E2+%2B+64x+%2B+36%29+=+0
%28x%2B2%29%282x%5E2+%2B+16x+%2B+9%29+=+0
Clearly no root of this equation is equal to 2. The roots are x+=+%28-16+%2B-+sqrt%28+16%5E2-4%2A2%2A9+%29%29%2F%282%2A2%29 & -2 i.e. x+=+-4+%2B-+sqrt%2846%29%2F2 & -2.
So all values of the x are negative. So the given equation is no longer valid in real domain.
Note that the equation is satisfied by x=2 if it is sqrt%28x%2B7%29+-+sqrt%28x%2B2%29=... and not sqrt%28x%2B7%29+%2B+sqrt%28x%2B2%29=...