Question

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by $60$. Assume the middle number to be $x$ and form a quadratic equation satisfying the above statement. Hence, find the three numbers.

Answer 1

Let the middle number be $\mathit{x}$ and hence, the other two numbers will be $\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}$ and $\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$

According to the question,

$$\begin{gathered} \Rightarrow {\left(x\right)}^2-[{(x+1)}^2-{(x-1)}^2]=60 \\ \Rightarrow x^2-[x^2+1+2x-x^2-1+2x]=60 \\ \Rightarrow x^2-4x-60=0 \\ \Rightarrow x^2+6x-10x-60=0 \\ \Rightarrow x(x+6)-10(x+6)=0 \\ \Rightarrow (x-10)(x+6)=0 \\ \Rightarrow x=10,-6 \end{gathered}$$

The number cannot be negative, so $x=10$.

Thus, the three numbers are $\mathbf{9}\mathbf{,}\mathbf{}\mathbf{10}\mathbf{,}\mathbf{}\mathbf{11}$.