Question

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If the polynomial p(x) = 6x^+4x + k... Has √5 and -√ 5 as zeros, find k.... Plz... Ans...

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Answer 1

Answer:

$p(x) = 6 {x}^{2} + 4x + k = 0$

$given \: zeroes \: are \: \sqrt{5} \: and \: \sqrt{ - 5}$

☆

$put \: x = \sqrt{5} \: in \: p(x)$

$p(x) = 6 ({ \sqrt{5} })^{2} + 4( \sqrt{5} ) + k = 0$

$30 + 4 \sqrt{5} + k = 0$

$k = - 4 \sqrt{5} - 30$

$k = -2 (2 \sqrt{5} + 15)$

☆

$put \: x = - \sqrt{5} \: in \: p(x)$

$p(x) = 6( - { \sqrt{5} })^{2} + 4( - \sqrt{5} ) + k = 0$

$30 - 4 \sqrt{5} + k = 0$

$k = 4 \sqrt{5} - 30$

$k = 2(2 \sqrt{5} - 15)$

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Answer 2

$\sf\huge\bold\pink{Answer:}$

$Answer: \\ p(x) = 6 {x}^{2} + 4x + k = 0 \\ p(x)=6x2+4x+k=0 \\ given \: zeroes \: are \: \sqrt{5} \: and \: \sqrt{ - 5} \\ given \: zeroes \: are \: 5 \: and \: −5 \\ \\ ☆ \\ \\ put \: x = \sqrt{5} \: in \: p(x) \: put \: x=5 \: in \: p(x) \\ p(x) = 6 ({ \sqrt{5} })^{2} + 4( \sqrt{5} ) + k = 0 \\ p(x)=6(5)2+4(5)+k=0 \\ 30 + 4 \sqrt{5} + k = 0 \\ 30+45+k=0 \\ k = - 4 \sqrt{5} - 30 \\ k=−45−30 \\ k = -2 (2 \sqrt{5} + 15) \\ k=−2(25+15) \\ \\ ☆$