Question

if alpha , bita are the roots of quadratic equation ax^2+bx+c = 0 , then write alpha^5+bita^5 in terms of a, b, c​

Answer 1

Answer:

Alpha⁵ + bita⁵ in terms of A,B,C

Step-by-step explanation:

Let us sort out one thing:

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

Having established this,

Using Factorisation,

${ \alpha }^{5} + { \beta }^{5} = ( \alpha + \beta )( { \alpha }^{4} - \alpha {}^{3} \beta + { \alpha {}^{2} }^{} { \beta }^{2} - { \alpha \beta } {}^{3} + { \beta }^{4}$

Now grouping the term of second bracket we get,

$( { \alpha }^{4} - { \alpha }^{3} \beta + { \alpha }^{2} { \beta }^{2} - { \alpha \beta }^{3} + { \beta }^{4} =$

$( { \alpha }^{4} + { \beta }^{4} + 2 { \alpha }^{2} { \beta }^{2} ) - { \alpha }^{2} { \beta }^{2} - \alpha \beta ( { \alpha }^{2} + { \beta }^{2} )$

Using the first established equation, the second bracket reduce to,

$( \frac{ {b}^{2} - 2ac }{ {a}^{2} } ) {}^{2} - (\frac{c}{a} ) {}^{2} - \frac{c}{a} \: . \: ( \frac{ {b}^{2} - 2ac }{ {a}^{2} } ) = \frac{ {b}^{4} + 5a {}^{2} {c - 5ab {}^{2} }c}{ {a}^{4} }$

Putting this value in place of the 2nd bracket, we get,

${ \alpha }^{5} + { \beta }^{5} = ( \alpha + \beta ) \: . \frac{ {b}^{4} + 5 {a}^{2} {c - 5 {ab}^{2} }c }{ {a}^{4} }$

$\frac{ - b}{a} \: . \: \frac{ {b}^{4} + 5 {a}^{2} {c}^{2} - 5 {ab}^{2}c }{ {a}^{4} }$

$= \frac{ - b( {b}^{4} + {5a}^{2} {c}^{2} - {5ab}^{2} c) }{ {a}^{5} }$