SOLUTION: Amother is 4 times older then her son 5 years ago the product of thier ages was 175 find thier present ages

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Question 840185: Amother is 4 times older then her son 5 years ago the product of thier ages was 175 find thier present ages
Answer by Alan3354(69443) About Me  (Show Source):
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Amother is 4 times older then her son 5 years ago the product of thier ages was 175 find thier present ages
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4 times older = 5 times as old.
M = 5S
(M-5)*(S-5) = 175
Sub for M
(5S-5)*(S-5) = 175
(S-1)*(S-5) = 35
S%5E2+-+6S+-+30+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B-30+=+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=%28-6%29%5E2-4%2A1%2A-30=156.

Discriminant d=156 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+156+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+156+%29%29%2F2%5C1+=+9.2449979983984
x%5B2%5D+=+%28-%28-6%29-sqrt%28+156+%29%29%2F2%5C1+=+-3.2449979983984

Quadratic expression 1x%5E2%2B-6x%2B-30 can be factored:
1x%5E2%2B-6x%2B-30+=+%28x-9.2449979983984%29%2A%28x--3.2449979983984%29
Again, the answer is: 9.2449979983984, -3.2449979983984. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B-30+%29

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The son is 9.245 years
Mom is 46.225 years
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The solution is not an integer because the author of the problem doesn't know the difference between 4 times as old and 4 times older.
Is 1 times older the same as 1 times as old?
Is 100% greater 1x as much? Or 2x?
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Don't ever mention "10 times smaller" since 1x smaller is zero.