SOLUTION: Find a and b such that the zeros of ax^2 + bx +24 are 3 and 4.

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Question 1014156: Find a and b such that the zeros of ax^2 + bx +24 are 3 and 4.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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equation with roots 3 and 4:
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%28x-3%29%28x-4%29=0
x%5E2-7x%2B12=0
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To make c=24, we must multiply by 2.
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2x%5E2-14x%2B24=0
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ANSWER: a=2 and b=-14
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CHECK:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-14x%2B24+=+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=%28-14%29%5E2-4%2A2%2A24=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--14%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-14%29%2Bsqrt%28+4+%29%29%2F2%5C2+=+4
x%5B2%5D+=+%28-%28-14%29-sqrt%28+4+%29%29%2F2%5C2+=+3

Quadratic expression 2x%5E2%2B-14x%2B24 can be factored:
2x%5E2%2B-14x%2B24+=+2%28x-4%29%2A%28x-3%29
Again, the answer is: 4, 3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-14%2Ax%2B24+%29