SOLUTION: A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 114. Find the present age of son.

Question 887591: A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 114. Find the present age of son.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = man's age
y = son's age.
x = 5*y

2 years ago the sum of the squares of their ages was equal to 114.
(x-2)^2 + (y-2)^2 = 114

simplify this to get:
x^2 - 4x + 4 + y^2 - 4y + 4 = 114

since x = 5y, replace x with 5y to get:
(5y)^2 - 4(5y) + 4 + y^2 - 4y + 4 = 114

simplify this to get:
25y^2 - 20y + 4 + y^2 - 4y + 4 = 114

combine like terms to get:

26y^2 - 24y + 8 = 114

subtract 114 from both sides of this equation to get:

26y^2 - 24y - 106 = 0

factor out the gcf of this equation to get:

2 * (13y^2 - 12y - 53) = 0

use the quadratic formula to factor (13y^2 - 12y - 53) to get:

y = 2.53275... or y = -1.60967...

since y can't be negative, then y has to be equal to 2.53275...

if i did this correctly, that's the age of his son presently.

that means the man is 5 * 2.53275... = 12.66377... years old.

that's not very old to be a father, but that's what the numbers show.

let's see if they're accurate in terms of the problem.

2 years ago, the sum of the squares of their ages was 114.

the man is 12.66377... years old
2 years ago, the man was 10.66377... years old.
square that and you get 113.71617...

the boy is 2.53275... years old.
2 years ago, the boy was 0.53275... years old.
square that and you get .283828...

add 113.71617... and .38382... and you get 114.

even though the ages don't look right, the formulas work out ok so i have to assume those are your answers.

the father's age is 12.66377... and the son's age is 2.53275...

you can round those number to whatever number of decimal places less than 5 that you want.

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