Question

Ans for if alpha and beta are the zeros of the polynomial x2+4x+3 from the polynomial whose zeros are 1+beta/alpha and 1+alpha/beta

Answer 1

Hey there!

The answer is in the image.

Hope it helps,
Purva
Brainly Community

Answer 2

$\sf \: {x}^{2} + 4x + 30$


$\sf \: {x}^{2} + x + 3x - 3 = 0$


$\sf \: x(x + 1) + 3(x + 1) = 0$


$\sf \: (x + 1)(x + 3) = 0$


$\underline{ \boxed{ \sf x = - 1, - 3}}$


$\sf \: 1 + \frac{ \beta}{ \alpha} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 + \frac{ \alpha}{ \beta}$


$\sf \alpha = - 1$


$\sf \beta = - 3$


$\sf \: 1 + \frac{ \cancel - 3}{ \cancel - 1}$


$\color{red} \sf \: = 4$


$\sf \: 1 + \frac{ - 1}{ - 3}$


$\sf \: 1 + \frac{1}{3} = \boxed{ \color{red} \frac{4}{3} }$


$\sf \: a,b$


$\sf \: {x}^{2} - (sum \: of \: roots)x + product = 0$


$\sf \: {x}^{2} - \bigg( \frac{16}{3} \bigg) x + \frac{16}{3} = 0$


$\color{blue} \sf \: {3x}^{2} - 16x + 16 = 0$