Question

The sum of a number and its reciprocal is $3\frac{41}{80}$. Find the number.

Answer 1

$$\begin{gathered} \text{Let the number be}x. \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}: \\ x+\frac{1}{x}=3\frac{41}{80} \\ \Rightarrow \frac{x^2+1}{x}=\frac{281}{80} \\ \Rightarrow 80x^2+80=281x \\ \Rightarrow 80x^2-281x+80=0 \\ \Rightarrow 80x^2-(256+25)x+80=0 \\ \Rightarrow 80x^2-256x-25x+80=0 \\ \Rightarrow 16x(5x-16)-5(5x-16)=0 \\ \Rightarrow (5x-16)(16x-5)=0 \\ \Rightarrow \text{5}x-16=0\text{or 16}x-5=0 \\ \Rightarrow x=\frac{16}{5}\text{or}x\text{=}\frac{5}{16} \\ \text{Hence, the number is either}\frac{16}{5}\text{or}\frac{5}{16}\text{.} \end{gathered}$$