Let number = x, then its reciprocal = 1/x
\(x+\frac{1}{x}=\frac{13}{6},(x^2+1)\times6=13x\)
6x2 - 13x + 6 = 0
\(x=\frac{13\pm \sqrt{(-13)^2-4\times6\times6}}{2\times6}\)
= \(\frac{13\pm \sqrt{169-144}}{12}=\frac{13\pm \sqrt{25}}{12}\)
Number = \(\frac{3}{2}\) or \(\frac{2}{3}\)
