SOLUTION: Please help me. sqrt ( x - 2 ) + 3 = sqrt ( 4x + 1 ) I know we're supposed to get one radical by itself, but one already is. And then the power rule to eliminate the radica

Algebra ->  Radicals -> SOLUTION: Please help me. sqrt ( x - 2 ) + 3 = sqrt ( 4x + 1 ) I know we're supposed to get one radical by itself, but one already is. And then the power rule to eliminate the radica      Log On


   

Question 349602: Please help me.
sqrt ( x - 2 ) + 3 = sqrt ( 4x + 1 )
I know we're supposed to get one radical by itself, but one already is. And then the power rule to eliminate the radicals. But from there, I'm stuck.
Thank you for your time.
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt+%28+x+-+2+%29+%2B+3+=+sqrt+%28+4x+%2B+1+%29
Square both sides.
%28x-2%29%2B6sqrt%28x-2%29%2B9=4x%2B1
6sqrt%28x-2%29=3x-6
Square both sides again,
36%28x-2%29=9x%5E2-36x%2B36
36x-72=9x%5E2-36x%2B36
9x%5E2-72x%2B108=0
x%5E2-8x%2B12=0
%28x-2%29%28x-6%29=0
Two solutions:
x-2=0
highlight%28x=2%29
.
.
.
x-6=0
highlight%28x=6%29

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
.
Yes! getting one radical by itself is the first step.
.
sqrt+%28+x+-+2+%29+%2B+3+=+sqrt+%28+4x+%2B+1+%29+
.
SQUARING BOTH SIDES OF THE EQUATION
%28x-2%29+%2B6+sqrt+%28+x+-+2+%29+%2B+9+=+4x+%2B+1++
.
isolating radical on one side of the equation once again
6+sqrt+%28+x+-+2+%29=+3x+-6++
3+sqrt+%28+x+-+2+%29=+x+-2+
.
SQUARING BOTH SIDES OF THE EQUATION
9%28x-2%29=+x%5E2+-+4x+%2B4+
.
Simplify and Solve or x
0=+x%5E2+-+13x+%2B22+
factor
(x-11)(x-2)=0
x = 2 or 11