Question

A year ago, the father was 8 times as old as his son. Now, his age is the square of his son's age. Find their present ages.

Answer 1

Step 1:

Form a quadratic equation:

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

Let the present age of father be $y$.

One year ago,

$\begin{array}{rcl}y-1& =& 8\left(x-1\right)\\ & \Rightarrow & y-1=8x-8\\ & \Rightarrow & y=8x-8+1\\ & \Rightarrow & y=8x-7\end{array}$

Now by given condition,

$\begin{array}{rcl}y& =& x^2\\ & & \end{array}$

$\begin{array}{rcl}So,\left(8x-7\right)& =& x^2\\ & \Rightarrow & x^2-8x+7=0\end{array}$

Step 2:

Solve the quadratic equation:

Use factorization method to solve the equation.

$\begin{array}{rcl}& \Rightarrow & x^2-1x-7x+7=0\\ & \Rightarrow & x\left(x-1\right)-7\left(x-1\right)=0\\ & \Rightarrow & \left(x-7\right)\left(x-1\right)=0\\ & \Rightarrow & x-7=0orx-1=0\\ & \Rightarrow & x=7orx=1\end{array}$

When $x=1$, the age of son is equal to the age of father, which is impossible, reject $x=1$

Take the value $x=7$

So, present age of son is $7\mathrm{years}$

And present age of father is $8\times 7-7=49\mathrm{years}$

Final answer:

Hence, the present age of son is $7\mathrm{years}$.

The present age of father is $49\mathrm{years}$.