Question

The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.

Answer 1

$$\begin{gathered} \text{Let the greater number be}x\text{and the smaller number be}y. \\ \text{According to the question:} \\ {x}^2-y^2=45...\left(\text{i}\right) \\ {y}^2=4x...\left(\mathrm{ii}\right) \\ \text{From}\left(\text{i}\right)\text{and}\left(\text{ii}\right),\text{we get:} \\ {x}^2-4x=45 \\ \Rightarrow x^2-4x-45=0 \\ \Rightarrow x^2-(9-5)x-45=0 \\ \Rightarrow x^2-9x+5x-45=0 \\ \Rightarrow x(x-9)+5(x-9)=0 \\ \Rightarrow (x-9)(x+5)=0 \\ \Rightarrow x-\text{9 = 0 or}x+5=0 \\ \Rightarrow x=9\text{or}x=-\text{5} \\ \Rightarrow x=9(\because x\text{is a natural number}) \\ \text{Putting the value of}x\text{in equation}\left(\text{ii}\right),\mathrm{we}\mathrm{get}: \\ {y}^2=4\times 9 \\ \Rightarrow y^2=36 \\ \Rightarrow y=6 \\ \text{Hence, the two numbers are 9 and 6}\text{.} \end{gathered}$$