```Question 204177
How do you solve this radical the {{{sqrt(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?
<pre><font size = 4 color = "indigo"><b>
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(2x+5)}}} you get just what is under the
radical {{{2x+5}}}, not {{{(2x+5)^2}}}. 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(2x+5)+5=x}}}

{{{sqrt(2x+5)=x-5}}}

{{{(sqrt(2x+5))^2=(x-5)^2}}}

{{{2x+5=(x-5)(x-5)}}}

{{{2x+5=x^2-5x-5x+25}}}

{{{2x+5=x^2-10x+25}}}

{{{0=x^2-12x+20}}}

{{{0=(x-10)(x-2)}}}

{{{x-10=0}}}, {{{x-2=0}}}

{{{x=10}}}, {{{x=2}}}

But we must check both solutions in the
original equations:

checking {{{x=10}}}
{{{sqrt(2x+5)+5=x}}}
{{{sqrt(2*10+5)+5=10}}}
{{{sqrt(20+5)+5=10}}}
{{{sqrt(25)+5=10}}}
{{{5+5=10}}}
{{{10=10}}}

That checks, so {{{x=10}}} is a solution:

checking {{{x=2}}}
{{{sqrt(2x+5)+5=x}}}
{{{sqrt(2*2+5)+5=2}}}
{{{sqrt(4+5)+5=2}}}
{{{sqrt(9)+5=2}}}
{{{3+5=2}}}
{{{8=2}}}

That doesn't check, so the only solution
is

{{{x=10}}}

Edwin</pre>
```