SOLUTION: 64.) Find all real solutions to each equation. {{{x^2+x+ sqrt(x^2+x)-2=0}}}

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Question 164961This question is from textbook Elementary and Intermediate
: 64.) Find all real solutions to each equation.
x%5E2%2Bx%2B+sqrt%28x%5E2%2Bx%29-2=0
This question is from textbook Elementary and Intermediate

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Put parentheses around the first two terms:

%28x%5E2%2Bx%29%2B+sqrt%28x%5E2%2Bx%29-2=0

Notice that %28x%5E2%2Bx%29 occurs in two places, so let the
variable part of the middle term, in this case the whole
middle term, sqrt%28x%5E2%2Bx%29be equal to the letter W.

That is to say we are letting 

sqrt%28x%5E2%2Bx%29+=+W

Now we square both sides:

%28sqrt%28x%5E2%2Bx%29%29%5E2+=+W%5E2

%28x%5E2%2Bx%29=+W%5E2

So we substitute W%5E2 for %28x%5E2%2Bx%29 in the original 
equation, and W for sqrt%28x%5E2%2Bx%29

%28x%5E2%2Bx%29%2B+sqrt%28x%5E2%2Bx%29-2=0

becomes the much simpler looking equation:

W%5E2%2BW-2=0

Factoring:

%28W%2B2%29%28W-1%29=0

Using the zero-factor property,

   

So now we have to substitute back.

W+=+sqrt%28x%5E2%2Bx%29



We can tell immediately that sqrt%28x%5E2%2Bx%29=-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%28x%5E2%2Bx%29=1

Square both sides:

%28sqrt%28x%5E2%2Bx%29%29%5E2=%281%29%5E2 

x%5E2%2Bx=1

x%5E2%2Bx-1=0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+

x+=+%28-1+%2B-+sqrt%28+1%2B4%29%29%2F%282%29+

x+=+%28-1+%2B-+sqrt%285%29%29%2F2+

Edwin