SOLUTION: (xsqr+6x)1/4as the sqr =2

Algebra ->  Radicals -> SOLUTION: (xsqr+6x)1/4as the sqr =2      Log On


   

Question 344472: (xsqr+6x)1/4as the sqr =2
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I can't decipher your question.
Use the caret to denote exponentiation, x^2=x%5E2
Use sqrt(x) to denote the square root of x, sqrt(x)=sqrt%28x%29
Is this what you mean??
root%284%2C%28x%5E2%2B6x%29%29=2
x%5E2%2B6x=2%5E4
x%5E2%2B6x=16
x%5E2%2B6x-16=0
%28x%2B8%29%28x-2%29=0
Two solutions:
x%2B8=0
highlight%28x=-8%29
.
.
x-2=0
highlight%28x=2%29