Question

if alpha and beta are zero of polynomial of 3x^2+2x–6then find the value of 1/alpha +1/beta​

Answer 1

The given polynomial is

$3 {x}^{2} + 2x - 6$

Comparing this polynomial with standard polynomial

$a {x}^{2} + bx + c$

We get

a = 3

b =2

c = - 6

Since

$\alpha \: \: \: and \: \beta$

are zeroes of the given polynomial

$3 {x}^{2} + 2x - 6$

Then,

$\alpha + \beta = - \frac{b}{a} = - \frac{2}{3}$

AND

$\alpha \beta = \frac{c}{a} = - \frac{6}{3} = - 2 \: \: \: (1)$

We know that

${( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ \\ = {( \alpha - \beta )}^{2} = - { (\frac{ 2}{3}) }^{2} - 4 \times - 2 \\ \\ = {( \alpha - \beta )}^{2} = \frac{4}{9} + 8 \\ \\ = {( \alpha - \beta )}^{2} = \frac{4 + 72}{9} = \frac{76}{9}$

THEREFORE

$\alpha - \beta = \frac{ \sqrt{76} }{3} \\ \\ = \frac{ \sqrt{4 \times 19} }{3} \\ \\ = \frac{2 \sqrt{19} }{3} \: \: \: (2)$

WE KNOW THAT

${( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta \\ \\ {( - \frac{2}{3} )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2( - 2) \\ \\ \frac{4}{9} = { \alpha }^{2} + { \beta }^{2} - 4 \\ \\ \frac{4}{9} + 4 = { \alpha }^{2} + { \beta }^{2} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{40}{9} \: \: \: \: (3)$

WE KNOW THAT

${ (\alpha + \beta ) }^{3} = { \alpha }^{3} + { \beta }^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ ( - { \frac{ 2}{3}) }^{3} = { \alpha }^{3} + { \beta }^{3} + 3( - 2)( - \frac{2}{3} ) \\ \\ - \frac{8}{27} = { \alpha }^{3} + { \beta }^{3} + 4 \\ \\ - \frac{8}{27} - 4 = { \alpha }^{ 3} + { \beta }^{3} \\ \\ \frac{ - 8 - 108}{27} = { \alpha }^{3} + { \beta }^{3} \\ \\ { \alpha }^{3} + { \beta }^{3} = - \frac{116}{27} \: \: \: (4)$

Then

$\frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ = \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ = \frac{ - \frac{ 2 }{3} }{ - \frac{ 2}{1} } \\ \\ = - \frac{2}{3} \times - \frac{1}{2} \\ \\ = \frac{1}{3}$

#BEBRAINLY