SOLUTION: √x+6 + √2-x = 4

Algebra ->  Square-cubic-other-roots -> SOLUTION: √x+6 + √2-x = 4      Log On


   

Question 298755: √x+6 + √2-x = 4
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt%28x%2B6%29%2Bsqrt%282-x%29=4

Pick one of the radicals and label it R.

If we pick the first radical to label R then we have
R=sqrt%28x%2B6%29=R.  So substituting R for the first
radical:

R%2Bsqrt%282-x%29=4

Isolate the radical term by subtracting R from both sides::

sqrt%282-x%29=4-R

Square both sides:

%28sqrt%282-x%29%29%5E2=%284-R%29%5E2

2-x=%284-R%29%284-R%29

FOIL out the right sides:

2-x=16-4R-4R%2BR%5E2

2-x=16-8R%2BR%5E2

Substitute sqrt%28x%2B6%29 for R

2-x=16-8sqrt%28x%2B6%29%2B%28sqrt%28x%2B6%29%29%5E2

2-x=16-8sqrt%28x%2B6%29%2Bx%2B6

Isolate the radical term and simplify the above:


8sqrt%28x%2B6%29=20%2B2x

To make things easier divide every term by 2

4sqrt%28x%2B6%29=10%2Bx

Square both sides:

%284sqrt%28x%2B6%29%29%5E2=%2810%2Bx%29%5E2

4%5E2%2A%28sqrt%28x%2B6%29%29%5E2=%2810%2Bx%29%2810%2Bx%29

16%28x%2B6%29=100%2B10x%2B10x%2Bx%5E2

16x%2B96=100%2B20x%2Bx%5E2

Get 0 on the left side:

0=x%5E2%2B4x%2B4

0=%28x%2B2%29%28x%2B2%29

0=%28x%2B2%29%5E2

x%2B2%29=0

x=-2

But we must ALWAYS check equations with even root
radicals as we often get extraneous solutions:

Checking:
sqrt%28x%2B6%29%2Bsqrt%282-x%29=4
sqrt%28%28-2%29%2B6%29%2Bsqrt%282-%28-2%29%29=4
sqrt%284%29%2Bsqrt%282%2B2%29=4
2%2Bsqrt%284%29=4
2%2B2=4
4=4

It checks, so the solution is x=-2 

Edwin<