Question

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

Answer 1

Let $x$ be the age of the son one year ago, then the age of the man was $8x$.

The present age of the son is $\left(x+1\right)$ and that of the man is $\left(8x+1\right)$. Then,

$8x+1={\left(x+1\right)}^2$

$8x+1=x^2+1+2x$

$x^2=6x$

$x=0,6$

As the age cannot be 0, so the value of $x$ is 6.

So, present age of son $=(x+1)=7$ years and present age of man $=(8x+1)=49$ years