Question 460650
<pre>
{{{ sqrt(3x+3)+sqrt(x+2)=5 }}}

Let u = one of the radical terms, say

{{{u=sqrt(x+2)}}}

{{{ sqrt(3x+3)+u=5 }}}

Isolate the radical term:

{{{ sqrt(3x+3)=5-u}}}

Square both sides:

{{{3x+3 = 25-10u+u^2}}}

Replace u by {{{sqrt(x+2)}}} and u² by x+2

{{{3x+3 = 25-10sqrt(x+2)+x+2}}}

{{{3x+3 = 27-10sqrt(x+2)+x}}}

Isolate the radical term:

{{{10sqrt(x+2)=24-2x}}}

Divide every term by 2 since they are all even:

{{{5sqrt(x+2)=12-x}}}

Square both sides:

{{{25(x+2) = 144-24x+x^2}}}

25x + 50 = 144 - 24x + x²

Get 0 on the left side:

0 = x² - 49x + 94

Factor the right side:

0 = (x - 2)(x - 47}

Use the zero factor principle

x - 2 = 0        x - 47 = 0
    x = 2             x = 47

But we must always check a radical equation,
because sometimes there may be a "phony"
solution, called "extraneous".  

Checking x = 2

{{{ sqrt(3x+3)+sqrt(x+2)=5 }}}
{{{ sqrt(3(2)+3)+sqrt(2+2)=5 }}}
{{{ sqrt(6+3)+sqrt(2+2)=5 }}}
{{{ sqrt(9)+sqrt(4)=5 }}}
{{{3+2=5}}}
{{{5=5}}}

That checks so x = 2 is a solution.

Checking x = 47

{{{ sqrt(3x+3)+sqrt(x+2)=5 }}}
{{{ sqrt(3(47)+3)+sqrt(47+2)=5 }}}
{{{ sqrt(141+3)+sqrt(49)=5 }}}
{{{ sqrt(144)+7=5 }}}
{{{12+7=5}}}
{{{19=5}}}

That is false so x = 47 is a "phony"
or "irrational" solution.  So the
only solution is x = 2

Edwin</pre>