Question

The sum of the squares of the two natural number is 34. If the first no. is one less than twice the 2^nd no. Find the numbers?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{second}\mathrm{number}=x \\ \mathrm{then}\mathrm{first}\mathrm{number}=2x-1 \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}, \\ {x}^2+{\left(2x-1\right)}^2=34 \\ \Rightarrow x^2+4x^2+1-4x=34 \\ \Rightarrow 5x^2-4x-33=0 \\ \Rightarrow 5x^2-15x+11x-33=0 \\ \Rightarrow 5x\left(x-3\right)+11\left(x-3\right)=0 \\ \Rightarrow \left(5x+11\right)\left(x-3\right)=0 \\ \Rightarrow 5x+11=0\mathrm{or}x-3=0 \\ \Rightarrow x=-\frac{11}{5}\left(\mathrm{rejected}\right)\mathrm{or}x=3 \\ \mathrm{So},\mathrm{second}\mathrm{number}=3 \\ \mathrm{first}\mathrm{number}=2\times 3-1=5 \end{gathered}$$