SOLUTION: How would you solve this equation? {{{ sqrt ( 2x ) = sqrt ( 3x+12 )-2 }}}

Algebra ->  Radicals -> SOLUTION: How would you solve this equation? {{{ sqrt ( 2x ) = sqrt ( 3x+12 )-2 }}}      Log On


   



Question 707329: How would you solve this equation?
+sqrt+%28+2x+%29+=+sqrt+%28+3x%2B12+%29-2+

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39790) About Me  (Show Source):
You can put this solution on YOUR website!
Square both sides and simplify. You may make to:
x%2B16=2%2Asqrt%283x%2B12%29

Square both sides again and simplify, getting:
x%5E2%2B20x%2B208=0

Easiest to use formula for solution to quadratic equation:
x=%28-20%2Bsqrt%2820%5E2-4%2A208%29%29%2F2
OR
x=%28-20-sqrt%2820%5E2-4%2A208%29%29%2F2
and find what you get.

You get -10+6i or -10-6i
(complex results)

Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
How would you solve this equation?

+sqrt+%28+2x+%29+=+sqrt+%28+3x%2B12+%29-2+
**************************************
The other person's response, that "You get -10+6i or -10-6i (complex results)", is WRONG11

sqrt%282x%29+=+sqrt%283x%2B12%29+-+2
As the smaller radicand, 2x MUST be greater than or equal to 0, we get: 2x+%3E=+0. So, x+%3E=+0.  
This gives us:      sqrt%282x%29+=+sqrt%283x%2B12%29+-+2, with x+%3E=+0
                         %28sqrt%282x%29%29%5E2+=+%28sqrt%283x+%2B+12%29+-+2%29%5E2 ---- Squaring both sides
                               2x+=+3x+%2B+12+-+4sqrt%283x+%2B+12%29+%2B+4
                    4sqrt%283x+%2B+12%29+=+3x+%2B+12+%2B+4+-+2x
                    4sqrt%283x+%2B+12%29+=+x+%2B+16
                  16%283x+%2B+12%29+=+x%5E2+%2B+32x+%2B+256 ----- Squaring both sides
                     48x+%2B+192+=+x%5E2+%2B+32x+%2B+256
x%5E2+%2B+32x+%2B+256+-+48x+-+192+=+0
                 x%5E2+-+16x+%2B+64+=+0
                        %28x+-+8%29%5E2+=+0
                          x - 8 = 0.
                               highlight%28x+=+8%29 <==== VALID, and ACCEPTABLE solution for x, in THIS case!!