Question

If $m$ and $n$ are the zeroes of the polynomial $3x^2+11x-4$, find the values of $\frac{m}{n}+\frac{n}{m}$.

Answer 1

$\frac{m}{n}+\frac{n}{m}$

$=\frac{m^2+n^2}{mn}$

$=\frac{{(m+n)}^2-2mn}{mn}$

$\Rightarrow \frac{m}{n}+\frac{n}{m}=\frac{{(m+n)}^2}{mn}-2.....\left(1\right)$

Given that $m$ and $n$ are the zeroes of the polynomial $3x^2+11x-4$.

Therefore,

Sum of zeroes $=m+n=-\frac{b}{a}=-\frac{11}{3}$

Product of zeroes $=mn=\frac{c}{a}=\frac{-4}{3}$

$\therefore \frac{m}{n}+\frac{n}{m}=\frac{{\left(\frac{-11}{3}\right)}^2}{\left(\frac{-4}{3}\right)}-2=-\frac{121}{16}-2=-\frac{153}{16}$

Hence the value of $\frac{m}{n}+\frac{n}{m}$ wil be $-\frac{153}{16}$.