Question

If alpha and beta are zeroes of 3x^2+11x-4 .Find the value of alpha÷beta+ beta÷alpha

Answer 1

Given p(x)= 3x^2+11x-4

we know that,

sum of zeroes= -b/a

$\alpha + \beta = \frac{ - 11}{3}$
______________(1)

Also,

sum of zeroes= c/a

$\alpha \beta = \frac{ - 4}{3}$
__________________(2)

now,


$= \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta }$

using identity,

${x}^{2} + {y}^{2} = (x + y)^{2} - 2xy$
we have,

$\frac{( \alpha + \beta )^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ from \: 1 \: and \: 2 \: \\ \\ = (\frac{ { - 11}^{2} }{3^{2} } - \frac{2 \times - 4}{3} ) \div \frac{ - 4}{3} \\ \\ =( \frac{ 121}{9} + \frac{8}{3} ) \div \frac{ - 4}{3} \\ \\ = \frac{145}{9} \times - \frac{3}{4} \\ \\ = - \frac{145}{12}$

_______________________


hope it helps!!