Question

If alpha ,beta are the roots of equation x^2-2√3x +4 , prove that alpha cud + deta cub =0

Answer 1

Answer:

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Answer 2

$given \: equation \\ on \: comparing \: \: \: \: \: {x}^{2} - 2 \sqrt{3} x + 4 \\ \: \: \: \: \: \: \: \: with \: \: \: \: \: \: a {x}^{2} \: + b x + c\\a = 1 \: \: \: b = - 2 \sqrt{3} \: \: \: \: \: c \: = \: 4 \\ \because \: roots \: are \: \alpha \: and \: \beta \\ ( \alpha + \beta ) = \frac{ - b}{a} \\ ( \alpha + \beta ) = \frac{ - ( - 2 \sqrt{3} )}{1} \\ ( \alpha + \beta ) = 2 \sqrt{3} \\ and \: \alpha \beta = \frac{c}{a} \\ \alpha \beta = \frac{4}{1} \\ \alpha \beta = 4 \\ now \\ ( \alpha + \beta ) = 2 \sqrt{3} \\ on \: cubing \: both \: sides \\ {( \alpha + \beta )}^{3} = {(2 \sqrt{3}) }^{3} \\ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta ) = 8 \times 3 \sqrt{3} \\ put \: values \: \\ { \alpha }^{3} + { \beta }^{3} + 3 \times 4 \times 2 \sqrt{3} = 24 \sqrt{3} \\ { \alpha }^{3} + { \beta }^{3} + 24 \sqrt{3} = 24 \sqrt{3} \\ { \alpha }^{3} + { \beta }^{3} = 24 \sqrt{3 } - 24 \sqrt{3} \\ { \alpha }^{3} + { \beta }^{3} = 0 \\ hence \: proved$

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