Question

M and n are the zeros of the polynomial 3x square + 11 x minus 4 find the value of m by n plus n bi m ​

Answer 1

REFER TO THE ATTACHMENT FOR THE ANSWER

Answer 2

$\bf{\underline{\underline{Given}}}$

$\mathsf{\star\:\: p(x) = 3x^{2} + 11x - 4}$

$\mathsf{\star\:\: m\: \:and\:\:n \: are\: the\:zeroes\:of\:p(x)}$

$\bf{\underline{\underline{To\:\:find}}}$

$\mathsf{\star\:\: Value\:of\:\: \frac{m}{n} + \frac{n}{m}}$

$\bf{\underline{\underline{Solution}}}$

Let m and n are the zeroes of the polynomial

We know that

$\mathsf{Sum\:of\:zeroes = \frac{- (x\:coefficient)}{x^{2} \:coefficient}}$

$\mathsf{\rightarrow\: m + n = \frac{-11}{3}}$

$\mathsf{Product\:of\:zeroes = \frac{constant\:term}{x^{2} \:coefficient}}$

$\mathsf{\rightarrow\: mn = \frac{- 4}{3}}$

Now,

$\mathtt{\implies\: \frac{m}{n} + \frac{n}{m}}$

$\mathtt{\implies\: \frac{m^{2} + n^{2}}{m × n}}$

$\mathtt{\implies\: \frac{(m + n)^{2} - 2mm}{mn}}$

$\large{\mathtt{\implies\: \frac{(\frac{-11}{3})^{2} - 2 × (\frac{-4}{3})}{\large{\frac{-4}{3}}}}}$

$\large{\mathtt{\implies\: \frac{\frac{121}{9} + \frac{8}{3}}{\frac{-4}{3}}}}$

$\large{\mathtt{\implies\: \frac{\frac{145}{9}}{\frac{-4}{3}}}}$

$\mathtt{\implies\: \frac{145}{\cancel{9}} × \frac{\cancel{-3}}{4}}$

$\mathtt{\implies\: \green{\frac{-145}{12}}}$

∴ Required value = (-145)/12