Question

If m and n are the zeroes of the polynomial 3x^2+11x-4, find value of m/n+n/m.

Answer 1

Solution:-

•Since m and n are the zeros of the polynomial 3x²+11x-4, therefore,

$\sf\large \: \: \: \: m + n = - \dfrac{11}{3} \ \ \ \ \ \ \:...(1)$

And

$\sf \large\ \ \ \ \ \: mn = \dfrac{ - 4}{3} \ \ \ \ \ \ \ \: ...(2)$

Now,

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \: \dfrac{m}{n} + \dfrac{n}{m} = \dfrac{ {m}^{2} + {n}^{2} }{mn}$

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \: = \dfrac{ {(m + n) - 2mn}^{2} }{mn}$

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \: = \dfrac{ \bigg( - \dfrac{11}{3} \bigg)^{2} - 2 \bigg( - \dfrac{4}{3} \bigg) }{ - \dfrac{4}{3} }$

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \: = \dfrac{ \dfrac{121}{9} + \dfrac{8}{3} }{ - \dfrac{4}{3} }$

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \: = - \dfrac{145}{9} \times \dfrac{3}{4}$

⠀⠀⠀⠀⠀⠀⠀

$\sf\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \: = - \dfrac{145}{12}$

⠀⠀⠀⠀⠀⠀⠀

Therefore , the required value is -145/12