Question

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

Answer 1

Hint:
Here we take a variable x to denote unknown ages and solve the problem.
Let x be the present age of son.
And the present age of man is \[2x^2\] years.
8 years hence: -
Age of the son = \[x{\rm{ }} + {\rm{ }}8\] years.
Age of the man = \[2x^2{\rm{ }} + {\rm{ }}8\] years.

Complete step by step solution:
⇒ \[\left( {{\bf{2x}}^2{\rm{ }} + {\rm{ }}{\bf{8}}} \right){\rm{ }} = {\rm{ }}{\bf{3}}\left( {{\bf{x}}{\rm{ }} + {\rm{ }}{\bf{8}}} \right){\rm{ }} + {\rm{ }}{\bf{4}}\]
⇒ \[{\bf{2x}}^2{\rm{ }} - {\rm{ }}{\bf{3x}}{\rm{ }} - {\rm{ }}{\bf{20}}{\rm{ }} = {\rm{ }}{\bf{0}}\]
⇒ \[2x^2{\rm{ }} - {\rm{ }}8x{\rm{ }} + {\rm{ }}5x{\rm{ }} - {\rm{ }}20{\rm{ }} = {\rm{ }}0\]
⇒ \[2x\left( {x{\rm{ }} - {\rm{ }}4} \right){\rm{ }} + {\rm{ }}5\left( {x{\rm{ }} - {\rm{ }}4} \right){\rm{ }} = {\rm{ }}0\] ⇒ (x - 4) (2x + 5) = 0
⇒ \[x{\rm{ }} - {\rm{ }}4{\rm{ }} = {\rm{ }}0{\rm{ }} or {\rm{ }}2x{\rm{ }} + {\rm{ }}5{\rm{ }} = {\rm{ }}0\]
⇒ \[{\bf{x}}{\rm{ }} = {\rm{ }}{\bf{4}},{\rm{ }} - {\rm{ }}{\bf{5}}/{\bf{2}}\;\] (As x can't be negative)
⇒ \[{\bf{x}}{\rm{ }} = {\rm{ }}{\bf{4}}\]
Son's present age = \[{\bf{x}}{\rm{ }} = {\rm{ }}{\bf{4}}\] years.
Man's present age = \[2x^2{\rm{ }} = {\rm{ }}2\left( 4 \right){\rm{^2 }} = {\rm{ }}2\left( {16} \right){\rm{ }} = \;{\bf{32}}{\rm{ }}{\bf{years}}\]
Hence, the present age of son and man is 4 years and 32 years.

Note:
If the present age is y, then n times the present age = ny. If the present age is x, then age, n years later/hence = x + n. If the present age is x, then age ,n years ago = x – n. The ages in a ratio a: b will be ax and bx. If the current age is y, then 1/n of the age is y/n.