Question

if alpha,beta are the zeroes of ax2+bx+c then find alpha5+beta5

Answer 1

Answer:

${ \alpha }^{5} \: + \: { \beta }^{5} = \dfrac{b( {b}^{2} - 2ac) \: (3ac - {b}^{2}) + b {a}^{2} {c}^{2} }{ {a}^{5} }$

Step-by-step explanation:

Given--->

$\alpha \: and \: \beta \: are \: zeroes \: of \: \\ a {x}^{2} + bx + c$

To find ---->

$value \: of \: \: ( \: { \alpha }^{5} + { \beta }^{5} \: )$

Concept used--->

1)

$if \: \: a {x}^{2} \: + bx \: + \: c \: is \: a \: quadratic \: polynomial \: then \\ sum \: of \: zeroes \: = - \dfrac{b}{a} \\ product \: of \: zeroes \: = \dfrac{c}{a}$

2)

$( { \alpha }^{2} + { \beta }^{2}) \: ( { \alpha }^{3} + { \beta }^{3}) = { \alpha }^{5} + { \alpha }^{2} { \beta }^{3} + { \beta }^{2} { \alpha }^{3} + { \beta }^{5} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \alpha }^{5} + { \beta }^{5} + { \alpha }^{2} { \beta }^{2}( \alpha + \beta ) \\ = > ( { \alpha }^{5} + { \beta }^{5}) = ( { \alpha }^{2} + { \beta }^{2})( { \alpha }^{3} + { \beta }^{3}) - { \alpha }^{2} { \beta }^{2}( \alpha + \beta )$

Solution---->

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

$so$

$\alpha + \beta = - \dfrac{b}{a}$

$\alpha \beta = \dfrac{c}{a}$

$now$

${ \alpha }^{2} + { \beta }^{2} = ({ \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta) - 2 \alpha \beta$

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta$

${ \alpha }^{2} + { \beta }^{2} = {( - \dfrac{b}{a}) }^{2} - 2 ( \: \dfrac{c}{a} \: )$

${ \alpha }^{2} + { \beta }^{2} = \dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{2c}{a}$

${ \alpha }^{2} + { \beta }^{2} = \dfrac{ {b}^{2} - 2ac }{ {a}^{2} }$

$now$

${ \alpha }^{3} + { \beta }^{3} = ( \alpha + \beta ) \: ( { \alpha }^{2} + { \beta }^{2} - \alpha \beta )$

$= ( \: - \dfrac{b}{a} \: ) \: ( \: \dfrac{ {b}^{2} - 2ac }{ {a}^{2} } - \dfrac{c}{a} \: )$

$= ( \: - \dfrac{b}{a} \: ) \: ( \: \dfrac{ {b}^{2} - 2ac - ac }{ {a}^{2} } \: )$

$= - \dfrac{b \: ( \: {b}^{2} - 3ac \: ) }{ {a}^{3} }$

$= \dfrac{b \: (3ac \: - {b}^{2} \: ) }{ {a}^{3} }$

$now$

$( \: { \alpha }^{5} \: + { \beta }^{5} \: )$

$= ( \: { \alpha }^{2} + { \beta }^{2} \: ) \: ( { \alpha }^{3} + { \beta }^{3} ) - {( \alpha \beta) }^{2} \: ( \: \alpha + \beta \: )$

$= ( \dfrac{ {b}^{2} - 2ac }{ {a}^{2} } ) \: \dfrac{b \: (3ac - {b}^{2}) }{ {a}^{3} } - {( \: \dfrac{c}{a} \: )}^{2} \: ( - \dfrac{b}{a} )$

$= \dfrac{b( {b}^{2} - 2ac) \: (3ac - {b}^{2}) }{ {a}^{5} } + \dfrac{b {c}^{2} }{ {a}^{3} }$

$= \dfrac{b \: ( {b}^{2} - 2ac) \: (3ac - {b}^{2}) + b {a}^{2} {c}^{2} }{ {a}^{5} }$