SOLUTION: A father's age is 3 more than four times the age of his son. If the product of their ages is 351, what is the father's age?

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Question 253251: A father's age is 3 more than four times the age of his son. If the product of their ages is 351, what is the father's age?
Found 2 solutions by Greenfinch, JimboP1977:
Answer by Greenfinch(383)   (Show Source): You can put this solution on YOUR website!
If the son is Z years old, the father is 4Z + 3 years old
Product is Z(4Z + 3) = 4Z^2 + 3Z = 351
So 4Z^2 + 3Z - 351 = 0
You can plug into the formula now or try factorisation
Factors of 351 are 3x3x3x13
3 x 13 gives 39 and 3 x 3 is 9, multiplied by the 4 from the z^2 term gives 36, difference 3 which is the Z term so
(4Z + 39)(Z - 9)=0
Z = -39 or 9
So son's age is 9, father's age is 39

Answer by JimboP1977(311)   (Show Source): You can put this solution on YOUR website!
Father's Age = F
Son's Age = S







Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=5625 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 9, -9.75. Here's your graph:


Son's age is 9, Father's age is 39




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