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

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

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

Let the present age of the son be $x$ years.

$\therefore$ Present age of the $\operatorname{man}=x^{2}$ years

One year ago,

Age of the son $=(x-1)$ years

Age of the man $=\left(x^{2}-1\right)$ years

According to the given condition,

Age of the man $=8 \times$ Age of the son

$\begin{array}{l} \therefore x^{2}-1=8(x-1) \\ \Rightarrow x^{2}-1=8 x-8 \\ \Rightarrow x^{2}-8 x+7=0 \\ \Rightarrow x^{2}-7 x-x+7=0 \\ \Rightarrow x(x-7)-1(x-7)=0 \\ \Rightarrow(x-1)(x-7)=0 \\ \Rightarrow x-1=0 \text { or } x-7=0 \\ \Rightarrow x=1 \text { or } x=7 \end{array}$