Question

if m and n are the zeros of the polynomial 3x^2+11x-4 find the value of m÷n+n÷m(specify the method)

Answer 1

The first method is that you can find the zeroes of the equation and then substitute the values.
$3 {x}^{2} + 11x - 4 = 0 \\ 3 {x}^{2} + 12x - x - 4 = 0 \\ 3x(x + 4) - 1(x + 4) = 0 \\ (3x - 1)(x + 4) = 0 \\ m = \frac{1}{3} \\ n = - 4 \\ \\ \frac{m}{n} + \frac{n}{m} = \frac{ - 1}{12 } - 12 = \frac{ - 1 - 144}{12} \\ = \frac{ - 145}{12}$
However, when the zeroes of the polynomial are complex or hard to work with, this method would work better.
$mn = \frac{ - 4}{3} \\ m + n = \frac{ - 11}{3} \\ {m}^{2} + {n}^{2} + 2mn = \frac{121}{9} \\ {m}^{2} + {n}^{2} = \frac{121}{9} + \frac{8}{3} \\ {m}^{2} + {n}^{2} = \frac{145}{9}$
Now consider this
$\frac{m}{n} + \frac{n}{m} = \frac{ {m}^{2} + {n}^{2} }{mn} \\ = \frac{ \frac{145}{9} }{ \frac{ - 4}{3} } = \frac{145 \times 3}{9 \times - 4} = \frac{ - 145}{12}$
Mark it as the brainliest answer!!