SOLUTION: I need to find the quadractic equation for 5 and 28

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Question 171760: I need to find the quadractic equation for 5 and 28
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I need to find the quadractic equation for 5 and 28
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x = 5 or x = 28
x-5 = 0
x -28 = 0
(x-5)*(x-28) = 0
x%5E2+-+33x+%2B+140+=+0
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Not necessary, but a check:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-33x%2B140+=+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=%28-33%29%5E2-4%2A1%2A140=529.

Discriminant d=529 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--33%2B-sqrt%28+529+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-33%29%2Bsqrt%28+529+%29%29%2F2%5C1+=+28
x%5B2%5D+=+%28-%28-33%29-sqrt%28+529+%29%29%2F2%5C1+=+5

Quadratic expression 1x%5E2%2B-33x%2B140 can be factored:
1x%5E2%2B-33x%2B140+=+%28x-28%29%2A%28x-5%29
Again, the answer is: 28, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-33%2Ax%2B140+%29