SOLUTION: Please help me solve this radical equation: √(y+10)=y-2

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Question 433003: Please help me solve this radical equation: √(y+10)=y-2
Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
square both sides
y+10=(y-2)^2
y+10 = y^2-4 y+4
-y^2+5 y+6 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case -1y%5E2%2B5y%2B6+=+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=%285%29%5E2-4%2A-1%2A6=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+49+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%285%29%2Bsqrt%28+49+%29%29%2F2%5C-1+=+-1
y%5B2%5D+=+%28-%285%29-sqrt%28+49+%29%29%2F2%5C-1+=+6

Quadratic expression -1y%5E2%2B5y%2B6 can be factored:
-1y%5E2%2B5y%2B6+=+-1%28y--1%29%2A%28y-6%29
Again, the answer is: -1, 6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-1%2Ax%5E2%2B5%2Ax%2B6+%29

y=-1 y=6