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

Question

Accepted Answer

Step 1: Let present age of son $= x$ years, present age of man $= 2x^2$ years
Step 2: ATQ: $2x^2 + 8 = 3(x + 8) + 4 \Rightarrow 2x^2 + 8 = 3x + 28$
Step 3: $2x^2 - 3x - 20 = 0$
Step 4: $(2x + 5)(x - 4) = 0 \Rightarrow x = -\frac{5}{2}$ (rejected) or $x = 4$
Step 5: Present age of son $= 4$ years, present age of man $= 2 	imes 16 = 32$ years

Question

Solution