Question 164961
<pre><font size=4 color="indigo"><b>
Put parentheses around the first two terms:

{{{(x^2+x)+ sqrt(x^2+x)-2=0}}}

Notice that {{{(x^2+x)}}} occurs in two places, so let the
variable part of the middle term, in this case the whole
middle term, {{{sqrt(x^2+x)}}}be equal to the letter {{{W}}}.

That is to say we are letting 

{{{sqrt(x^2+x) = W}}}

Now we square both sides:

{{{(sqrt(x^2+x))^2 = W^2}}}

{{{(x^2+x)= W^2}}}

So we substitute {{{W^2}}} for {{{(x^2+x)}}} in the original 
equation, and {{{W}}} for {{{sqrt(x^2+x)}}}

{{{(x^2+x)+ sqrt(x^2+x)-2=0}}}

becomes the much simpler looking equation:

{{{W^2+W-2=0}}}

Factoring:

{{{(W+2)(W-1)=0}}}

Using the zero-factor property,

{{{matrix(2,3,   W+2=0, ",", W-1=0,
              W=-2, ",",   W=1   )}}}   

So now we have to substitute back.

{{{W = sqrt(x^2+x)}}}

{{{matrix(2,3,   
W=-2,
 ",", 
  W=1,
   sqrt(x^2+x)=-2, 
",",
 sqrt(x^2+x)=1
    )
}}}

We can tell immediately that {{{sqrt(x^2+x)=-2}}} has no
solution because square roots when written as a radical
are never negative, so that square root on the left
could not equal to -2 on the right.

So we solve

{{{sqrt(x^2+x)=1}}}

Square both sides:

{{{(sqrt(x^2+x))^2=(1)^2}}} 

{{{x^2+x=1}}}

{{{x^2+x-1=0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-1 +- sqrt( 1^2-4*1*(-1) ))/(2*1) }}}

{{{x = (-1 +- sqrt( 1+4))/(2) }}}

{{{x = (-1 +- sqrt(5))/2 }}}

Edwin</pre>