SOLUTION: How do you solve this radical the square root of 2x+5+5=x? What I have tried: you isolate the radical to (2x+5)^2 =( x+5)^2 Then I got 5x+25=50 simplify to 5x+50/5 which gave m

Algebra ->  Radicals -> SOLUTION: How do you solve this radical the square root of 2x+5+5=x? What I have tried: you isolate the radical to (2x+5)^2 =( x+5)^2 Then I got 5x+25=50 simplify to 5x+50/5 which gave m      Log On


   

Question 204177: How do you solve this radical the square root of 2x+5+5=x?
What I have tried: you isolate the radical to (2x+5)^2 =( x+5)^2
Then I got 5x+25=50 simplify to 5x+50/5 which gave me a final answer of x=5. Am I on the right track?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve this radical the sqrt%282x%2B5%29%2B5=x?
What I have tried: you isolate the radical to (2x+5)^2 =( x+5)^2
Then I got 5x+25=50 simplify to 5x+50/5 which gave me a final answer of x=5. Am I on the right track?

No. Your first error was a sign error. You got x+5 on the right 
and it should have been x-5.

Your second error was in 'squaring a square root'.  When
you square sqrt%282x%2B5%29 you get just what is under the
radical 2x%2B5, not %282x%2B5%29%5E2. You don't square what's
under the radical! The squaring takes away BOTH the radical AND 
the square, and you get just what's under the radical in the
fourth step.

sqrt%282x%2B5%29%2B5=x

sqrt%282x%2B5%29=x-5

%28sqrt%282x%2B5%29%29%5E2=%28x-5%29%5E2

2x%2B5=%28x-5%29%28x-5%29

2x%2B5=x%5E2-5x-5x%2B25

2x%2B5=x%5E2-10x%2B25

0=x%5E2-12x%2B20

0=%28x-10%29%28x-2%29

x-10=0, x-2=0

x=10, x=2

But we must check both solutions in the 
original equations:

checking x=10
sqrt%282x%2B5%29%2B5=x
sqrt%282%2A10%2B5%29%2B5=10
sqrt%2820%2B5%29%2B5=10
sqrt%2825%29%2B5=10
5%2B5=10
10=10

That checks, so x=10 is a solution:

checking x=2
sqrt%282x%2B5%29%2B5=x
sqrt%282%2A2%2B5%29%2B5=2
sqrt%284%2B5%29%2B5=2
sqrt%289%29%2B5=2
3%2B5=2
8=2

That doesn't check, so the only solution
is

x=10

Edwin