Question

The sum of a number and it's reciprocal is 122/11 .find the number.
Explain it.

Answer 1

$\large \textsc{} \texttt{Let the number = x} \\ \\ \texttt{Then it's reciprocal = } \displaystyle \mathtt{\frac{1}{x}}$

$\displaystyle x+\frac{1}{x}=\frac{122}{11} \\ \\ \\ \frac{x^2}{x}+\frac{1}{x}=\frac{122}{11} \\ \\ \\ \frac{x^2+1}{x}=\frac{122}{11} \\ \\ \\ (x^2+1)11=122x \\ \\ 11x^2+11=122x \\ \\ 11x^2+11-122x=0 \\ \\ 11x^2-122x+11=0 \\ \\ 11x^2-121x-x+11=0 \\ \\ 11x(x-11)-(x-11)=0 \\ \\ (x-11)(11x-1)=0 \\ \\ \\ \\ \boxed{\ \ \ \ \ \huge \textit{x = 11\ \ \ ;\ \ \ x = }\huge \mathit{\frac{1}{11}}\ \ \ }$

$\left[\mathfrak{Note:}\ \displaystyle \frac{1}{11} \ \ \text{can also be the answer because it's not mentioned 'natural number' in the question}\right]$