The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.
Answers
Answer:
the solve to your problem.....
Answer:
Step-by-step explanation:
Let the present age of the son be x.
Age of Father = 2(x^2)
After 8 years,
Age of son = x + 8
Age of Father = 2(x^2) + 8 - (1)
Also, age of father = 3(x + 8) + 4 - (2)
From (1) and (2) -
2x^2 + 8 = 3(x + 8) + 4
= 3x + 24 + 4
= 3x + 28
2x^2 - 3x + 8 - 28 = 0
2x^2 - 3x - 20 = 0
2x^2 - 8x + 5x - 20 = 0
2x(x - 4) + 5(x - 4) = 0
(2x + 5)(x - 4) = 0
2x + 5 = 0 OR x - 4 = 0
x = -5/4(Not Possible) x = 4
Therefore,
Present age of son = 4 years
Present age of father = 2(4)^2
= 2(16)
= 32 years