Question

If 2 ±root 3 are the two zeros of a polynomial then which of the polynomial you can make with these roots?​

Answer 1

refer to the attachment about please mark me as brainliest

Answer 2

$Let \: \alpha = 2 + \sqrt{3} \: and \: \beta = 2-\sqrt{3} \\are \: two \: zeroes \: of \: a\\polynomial \: p(x)$

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$We \:know \: that ,$

$\blue { If \: \alpha \:and \: \beta \: are \: two }$

$\blue{ zeroes \: of \: a \: Quadratic \: polynomial }$

$\blue{p(x) , \:then \: p(x) = k[x^{2} - (\alpha+\beta)x+\alpha \beta ] }$

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$i) Sum \:of \:the \: zeroes = 2 + \sqrt{3} + 2 - \sqrt{3}$

$\implies \alpha + \beta = 4 \: --(1)$

$ii) product \:of \:the \: zeroes =( 2 + \sqrt{3}) ( 2 - \sqrt{3})$

$\implies \alpha \beta = 2^{2} - (\sqrt{3})^{2} \\= 4 - 3 \\= 1 \: --(2)$

Therefore.,

$\red{ Required \: polynomial } \\= k[x^{2} - 2x + 1 ]$

$If\: k = 1 , then \\\green {the \: polynomial \: is \:x^{2} - 2x + 3}$

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