Question

One year ago, a man was $8$ times as old as his son. Now his age is equal to the square of his sons age. Find the sum of their present ages.

Answer 1

$\Rightarrow$ Let present age of son be $x$.

$\Rightarrow$ Then, present age of father will be $x^2.$

$\Rightarrow$ One year ago age of son $=x-1$

$\Rightarrow$ One year ago age of father $=x^2-1$ ----- ( 1 )

$\Rightarrow$ One year ago given age of father $=8(x-1)$ ------ ( 2 )

From ( 1 ) and ( 2 )

$\Rightarrow$ $x^2-1=8(x-1)$

$\Rightarrow$ $x^2-1=8x-8$

$\Rightarrow$ $x^2-8x-1+8=0$

$\Rightarrow$ $x^2-8x+7=0$

$\Rightarrow$ $x^2-7x-x+7=0$

$\Rightarrow$ $x(x-7)-1(x-7)=0$

$\Rightarrow$ $(x-7)(x-1)=0$

$\Rightarrow$ $x=7$ and $x=1$

$\Rightarrow$ Age of son $\left(x\right)=7years$

$\Rightarrow$ Age of father $\left(x^2\right)=7^2=49years$

$\Rightarrow$ Sum of ages of son and father $=7+49=56years$