Question

if alpha, beta are the zeros of x^2+px+a find the value of (alpha/beta+2) (beta/alpha+2)

Answer 1

Given:

${x}^{2} + px + a$
here,

→ a= 1 , b= p and c= a

we know that,

sum of zeroes = -b/a

$\alpha + \beta = - p$

____________________(1)

Also,

product of zeroes= c/a

$\alpha \times \beta = a$
______________________(2)

now, we have to find the value of :

$( \frac{ \alpha }{ \beta } + 2)( \frac{ \beta }{ \alpha } + 2) \\ \\ = ( \frac{ \alpha + 2 \beta }{ \beta })( \frac{ \beta + 2 \alpha }{ \alpha } ) \\ \\ = \frac{( \alpha + 2 \beta )( \beta + 2 \alpha )}{ \beta \times \alpha } \\ \\ = \frac{ \alpha \beta + 2 { \alpha }^{2} + 2 { \beta }^{2} + 4 \beta \alpha }{ \alpha \beta } \\ \\ = \frac{ \alpha \beta + 2( { \alpha }^{2} + { \beta }^{2} )+ 4 \beta \alpha }{ \alpha \beta }$

using identity ,

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

$\frac{ \alpha \beta + 2[( \alpha + \beta )^{2}- 2 \alpha \beta] + 4 \beta \alpha }{ \alpha \beta }$
from (1) & (2) , we have,

$= \frac{ - p + 2[( - p)^{2} - 2a] + 4a }{a } \\ \\ = \frac{ - p + 2( {p}^{2} - 2a) + 4a}{a} \\ \\ = \frac{ - p + 2 {p}^{2} - 4a + 4a }{a} \\ \\ = \frac{2 {p}^{2} - p }{a} \\ \\ = \frac{p(2p - 1)}{a}$
__________________________