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
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Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ \alpha {}^{5} + \beta {}^{5} \\ \\ so \: given \: quadratic \: equation \: is \\ ax {}^{2} + bx + c = 0 \\ \\ let \: its \: two \: roots \: be \: \alpha \: and \: \beta \\ we \: know \: that \\ \\ \alpha + \beta = \frac{ - b}{a} \: \: \: \: \: \: \: (1) \\ \\ \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2) \\ \\ applying \: now \: \: (2) {}^{2} \\ \\ we \: get \\ \\ \alpha {}^{2} \beta {}^{2} = \frac{c {}^{2} }{a {}^{2} }$

$so \:we \: know \: that \\ \alpha {}^{2} + \beta {}^{2} = ( \alpha + \beta ) {}^{2} - 2 \alpha \beta \\ \\ = ( \frac{ - b}{a} \: ) {}^{2} - 2 \times \frac{c}{a} \: \: \: \: \: \: from \: (1) \: and \: (2) \\ \\ = \frac{b {}^{2} }{a {}^{2} } - \frac{2c}{a} \: \: \: \: \: \: \: (4)$

$similarly \\ \alpha {}^{3} + \beta {}^{3} = ( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ = ( \frac{ - b}{a} ) {}^{3} - 3 \times \frac{c}{a} \times ( \frac{ - b}{a} ) \\ \\ = \frac{ - b {}^{3} }{a {}^{3} } + \frac{3bc}{a {}^{2} } \: \: \: \: \: \: (5)$

$so \: here \\ \alpha {}^{5} + \beta {}^{5} = ( \alpha {}^{3} + \beta {}^{3} )( \alpha {}^{2} + \beta {}^{2} ) - \alpha {}^{2} \beta {}^{2} ( \alpha + \beta ) \\ \\ from \: (1) \: , \: (2) \: , \: (3),(4) \: and \: (5)\\ we \: get \\ \\ = ( \frac{ - b {}^{3} }{a {}^{3} } + \frac{3bc}{a {}^{2} } )( \frac{b {}^{2} }{a {}^{2} } - \frac{2c}{a} ) - \frac{c {}^{2} }{a {}^{2} } ( \frac{ - b}{a} ) \\ \\ = \frac{ - b {}^{5} }{a {}^{5} } + \frac{3b {}^{3} c}{a {}^{3} } \: + \frac{2b {}^{3}c }{a {}^{4} } \: - \frac{3bc {}^{2} }{a {}^{3} } + \frac{bc {}^{2} }{a {}^{3} } \\ \\ \\ = \frac{ - b {}^{5} }{a {}^{5} } + \frac{3b {}^{3}c }{a {}^{3} } + \frac{2b {}^{3} c}{a {}^{4} } - \frac{2bc {}^{2} }{a {}^{3} }$