Question

solve this question anyone​

Answer 1

Answer:

-145/12

Step-by-step explanation:

see attachment

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Answer 2

Given :

• m and n are the zeroes of the polynomial 3x² + 11x - 4 = 0

To Find :

• Value of $\sf{\dfrac{m}{n}+\dfrac{n}{m}}$

Solution :

We have given two zeroes of the polynomial. First consider the case of m and then case of n.

In the polynomial, instead of x replace m.

$\red{\implies}$ $\sf{3x^2+11x-4=0}$

$\red{\implies}$ $\sf{3m^2+11m-4=0}$

$\red{\implies}$ $\sf{3m^2\:+\:12m-1m-4=0}$

$\red{\implies}$ $\sf{3m(m+4)-1(m+4)=0}$

$\red{\implies}$ $\sf{(m+4)\:(3m-1) =0}$

$\red{\implies}$ $\sf{m+4=0\:\:or\:\:3m-1=0}$

$\red{\implies}$ $\sf{m=-4\:\:or\:\:3m=1}$

$\sf{m=-4\:or\:m=\dfrac{1}{3}}$

$\large{\boxed{\sf{\red{m\:=\:-4\:or\:\dfrac{1}{3}}}}}$

Now, replace x with n in the polynomial, 3x² + 11x - 4 = 0.

$\red{\implies}$ $\sf{3n^2\:+\:11n-4=0}$

$\red{\implies}$ $\sf{3n^2\:+\:12n-n-4=0}$

$\red{\implies}$ $\sf{3n(n+4)-1(n+4)=0}$

$\red{\implies}$ $\sf{(n+4)\:\:(3n-1)\:=0}$

$\red{\implies}$ $\sf{n+4=0\:\:or\:3n=1}$

$\red{\implies}$ $\sf{n=-4\:\:or\:\:3n=1}$

$\red{\implies}$ $\sf{n=-4\:or\:n=\dfrac{1}{3}}$

$\large{\boxed{\sf{\purple{n\:=\:-4\:or\:\dfrac{1}{3}}}}}$

Now, we have the value of m and n.

And we have to find, m/n + n/m.

Substitute, m and n = - 4, 1/3,

$\sf{\dfrac{m}{n}\:+\:\dfrac{n}{m}}$

$\red{\implies}$ $\frac{ \frac{1}{3} }{ -4} + \frac{ \frac{ - 4}{1} }{3}$

$\red{\implies}$ $\frac{1}{3} \times \frac{1 }{ - 4} +( - 4) \times \frac{3}{1}$

$\red{\implies}$ $\frac{1}{ - 12} + (- 12)$

$\red{\implies}$ $\frac{1}{ - 12} - 12$

$\red{\implies}$ $\frac{ - 1 - (144)}{12}$

$\red{\implies}$ $\frac{ - 1 - 144}{12}$

$\red{\implies}$ $\frac{ - 145}{12}$

$\large{\boxed{\sf{\purple{Value\:of\:\dfrac{m}{n}+\dfrac{n}{m}\:=\:\dfrac{-145}{12}}}}}$