SOLUTION: Solve the following quadratic equations: P^2+6p=0

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Question 333133: Solve the following quadratic equations: P^2+6p=0
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
p%5E2+%2B+6p=0
p%28p%2B6%29+=+0
so p=0 or p+6=0
thus p=0 or p =-6
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B0+=+0) has the following solutons:

x%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=%286%29%5E2-4%2A1%2A0=36.

Discriminant d=36 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+36+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+36+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%286%29-sqrt%28+36+%29%29%2F2%5C1+=+-6

Quadratic expression 1x%5E2%2B6x%2B0 can be factored:
1x%5E2%2B6x%2B0+=+1%28x-0%29%2A%28x--6%29
Again, the answer is: 0, -6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B0+%29