Question

The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be $4$ years more than three times the age of her son. Find their present age.

Answer 1

Step 1: Assuming the son's and mother's age in terms of variable

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

The age of the mother is twice the square of her son

So, the mother’s age be $2x^2$

Eight years hence,

Age of son $=(x+8)yrs$

Age of Mother $=(2x^2+8)yrs$

Step 2: Finding the value of $x$

According to the questions,

After $8$ years, the age of the mother will be $4$ years more than three times the age of her son

$$\begin{gathered} \Rightarrow 2x^2+8=3(x+8)+4 \\ \Rightarrow 2x^2+8=3x+24+4 \\ \Rightarrow 2x^2+8=3x+28 \\ \Rightarrow 2x^2-3x-20=0 \\ \Rightarrow 2x^2-8x+5x-20=0 \\ \Rightarrow 2x(x-4)+5(x-4)=0 \\ \Rightarrow (2x+5)(x-4)=0 \\ \Rightarrow x=\frac{-5}{4},x=4 \end{gathered}$$

The age of a person cannot be negative thus, $x=4$

Step 3: Finding the age of the son and mother

The age of the son is $4$ years

The age of the mother is $2x^2=2{\left(4\right)}^2=2\times 16=32yrs$

Hence, the present age of the son is $4$ years and his mother's present age is $32$ years.