SOLUTION: Please help. Find the real solution to: {{{x+sqrt(x)-6=0}}} Answer: x=4 But how?

Algebra ->  sets and operations  -> Lessons -> SOLUTION: Please help. Find the real solution to: {{{x+sqrt(x)-6=0}}} Answer: x=4 But how?       Log On


   

Question 981941: Please help. Find the real solution to:
x%2Bsqrt%28x%29-6=0
Answer: x=4
But how?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First isolate sqrt%28x%29


x%2Bsqrt%28x%29-6=0


x%2Bsqrt%28x%29-6%2B6=0%2B6


x%2Bsqrt%28x%29=6


sqrt%28x%29=6-x


Now square both sides to get rid of the square root.


sqrt%28x%29=6-x


%28sqrt%28x%29%29%5E2=%286-x%29%5E2


x=%286-x%29%5E2


x=36-12x%2Bx%5E2


x-x=36-12x%2Bx%5E2-x


0=36-13x%2Bx%5E2


x%5E2-13x%2B36=0


%28x-9%29%28x-4%29=0


x-9=0 or x-4=0


x=9 or x=4


If you check both possible solutions back in the original equation, only x+=+4 works. The other value x=9 does NOT satisfy the original equation. So it's not a true solution.