SOLUTION: A man is three times as old as his son. Ten years ago the sum of the square of their ages was 1250. find their ages

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: A man is three times as old as his son. Ten years ago the sum of the square of their ages was 1250. find their ages       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 956947: A man is three times as old as his son. Ten years ago the sum of the square of their ages was 1250. find their ages
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
M=man's age; S=son's age
M=3S
%28M-10%29%5E2%2B%28S-10%29%5E2=1250 Substitute for M
%283S-10%29%5E2%2B%28S-10%29%5E2=1250
9S%5E2-60S%2B100%2BS%5E2-20S%2B100=1250
10S%5E2-80S%2B200=1250 Subtract 1250 from each side.
10S%5E2-80S-1050=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aS%5E2%2BbS%2Bc=0 (in our case 10S%5E2%2B-80S%2B-1050+=+0) has the following solutons:

S%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=%28-80%29%5E2-4%2A10%2A-1050=48400.

Discriminant d=48400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--80%2B-sqrt%28+48400+%29%29%2F2%5Ca.

S%5B1%5D+=+%28-%28-80%29%2Bsqrt%28+48400+%29%29%2F2%5C10+=+15
S%5B2%5D+=+%28-%28-80%29-sqrt%28+48400+%29%29%2F2%5C10+=+-7

Quadratic expression 10S%5E2%2B-80S%2B-1050 can be factored:
10S%5E2%2B-80S%2B-1050+=+10%28S-15%29%2A%28S--7%29
Again, the answer is: 15, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+10%2Ax%5E2%2B-80%2Ax%2B-1050+%29

S=15 ANSWER The son is 15 years old.
M=3S=3(15yrs)=45yrs ANSWER 2: The man is 45 years old.
CHECK:
%28M-10%29%5E2%2B%28S-10%29%5E2=1250
%2845-10%29%5E2%2B%2815-10%29%5E2=1250
35%5E2%2B5%5E2=1250
1225%2B25=1250
1250=1250