SOLUTION: find real solutions. {{{x+sqrt(x)=6}}} Thanks Caroline

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Question 5225: find real solutions.
x%2Bsqrt%28x%29=6
Thanks Caroline
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
Caroline,
Try using y instead of sqrt%28x%29. x is sqrt%28x%29 squared, or y%5E2. So, your original equation becomes
y%5E2%2By-6+=+0. I will use my pluggable quadratic solver now:

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 1y%5E2%2B1y%2B-6+=+0) has the following solutons:

y%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+2
y%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C1+=+-3

Quadratic expression 1y%5E2%2B1y%2B-6 can be factored:
1y%5E2%2B1y%2B-6+=+1%28y-2%29%2A%28y--3%29
Again, the answer is: 2, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-6+%29


since y is sqrt(x), y cannot be negative, so solution -3 is invalid. You have y=2, y is sqrt%28x%29, so x is 4.