SOLUTION: solve. w^2-5w-50=0

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Question 331192: solve. w^2-5w-50=0
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
You can factor this or use the quadratic equation
w%5E2-5w+-+50+=+0
%28w-10%29%28w%2B5%29+=+0
w = 10 or w=-5
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-50+=+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=%28-5%29%5E2-4%2A1%2A-50=225.

Discriminant d=225 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+225+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+225+%29%29%2F2%5C1+=+10
x%5B2%5D+=+%28-%28-5%29-sqrt%28+225+%29%29%2F2%5C1+=+-5

Quadratic expression 1x%5E2%2B-5x%2B-50 can be factored:
1x%5E2%2B-5x%2B-50+=+1%28x-10%29%2A%28x--5%29
Again, the answer is: 10, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-50+%29