SOLUTION: The age of a man is twice the square of the age of his son.8 years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.

Click here to see ALL problems on Quadratic Equations

Question 866821: The age of a man is twice the square of the age of his son.8 years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.
Answer by dkppathak(439) (Show Source):
You can put this solution on YOUR website!
The age of a man is twice the square of the age of his son.8 years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.
Let age of son is X yrs
age of man will be 2X^2
8 yrs hence age of son will be= X+8 yrs
8 yrs hence age of man will be =2X^2+8
as per given conditions
2X^2+8= 3(X+8)+4
X^2 +8 =3x+24+4
2x^2+8 =3x+28
2X^2-3x-20=0
2X^2-8X+5X-20=0
2x(x-4)+5(x-4)=0
(x-4)(2x+5)=0
x=4 or x=-5/2
neglecting negative value of age
x=4
present age of son is 4 yrs
present age of man is 32 yrs