Question

The sum of a rational number \(x\) and its reciprocal is \(\frac{113}{56}\). If \(x=\frac{m}{n}\), where \(\gcd(m,n)=1\), then \(m^2+n^2=\ ?\)

• A. \(56\)
• B. \(113\)
• C. \(169\)
• D. \(57\)

Hint:

For \(x=\frac{m}{n}\), \[ x+\frac1x=\frac{m^2+n^2}{mn}. \] Compare numerator and denominator directly when the fraction is already in lowest terms.

Answer 1

The Correct Option is B

Given \[ x+\frac1x=\frac{113}{56} \] Let \[ x=\frac{m}{n} \] Then \[ \frac{m}{n}+\frac{n}{m} = \frac{m^2+n^2}{mn} = \frac{113}{56} \]

Step 1: Compare the fractions. Since \[ \gcd(113,56)=1 \] and \(\gcd(m,n)=1\), we obtain \[ m^2+n^2=113 \] and \[ mn=56 \] Therefore, \[ \boxed{113} \]