Question

Given that alpha and beta are the roots of the equation x2=7x+4
Show that alpha3=53alpha+28


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Answer 1

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$alpha \: is \: a \: root \: of \: the \: eqn \\ {x}^{2} = 7x + 4 \\ therefore \\ { \alpha }^{2} = 7 \alpha + 4 \: \: \: \: \: \: \: \: \: \: \: \: \: (1)\\multiply \: both \: side \: by \: \alpha \: of \: eqn(1) \\ { \alpha }^{3} = 7 { \alpha }^{2} \: + 4 \alpha \: \: \: \: \: \: \: \: \: \: \: (2) \\ subsitute \: the \: (1) \: into \: (2) \\ { \alpha }^{3} = 7(7 \alpha + 4 ) + 4 \alpha \\ = 49 \alpha + 28 + 4 \alpha \\= 53 \alpha + 28 \\ hence \: proved$

Answer 2

Answer:

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