Question

What number when multiplied by itself is $23$ greater than the preceding number when it is multiplied by itself?

Answer 1

Find the number:

Let the number be $x$.

Given that, number when multiplied by itself is $23$ greater than the preceding number when it is multiplied by itself

According to the given information,

$$\begin{gathered} x·x=23+\left(x-1\right)·\left(x-1\right) \\ {x}^2=23+{\left(x-1\right)}^2 \\ \Rightarrow x^2=23+x^2+1-2x\left[\because {\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ \Rightarrow x^2=24+x^2-2x \\ \Rightarrow x^2-x^2+2x=24\left[\mathrm{Subtracting}\left(x^2-2x\right)\mathrm{from}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow 2x=24 \\ \Rightarrow x=\frac{24}{2}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2\right] \\ \therefore x=12 \end{gathered}$$

Hence, the required number is $12$.