Question

if alpha and beta are the two zeroes of the polynomial xsquare-3x+1 then find a quadratic polynomial whose zeroes are 1/2alpha square and 1/2 beta square

Q 5 OF THE PDF.KINDLY HELP ONLY IF YOU KNOW THE ANSWER..

0cae77802640dd357b5b7b98489e4dcc.pdf

Answer 1

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See The Attachment My Friend.....

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Answer 2

$Let \: p(x) = x^{2} + 4x + 3$

$Compare \: x^{2} + 4x + 3 \:with \:ax^{2} +bx+c,\\we \:get$

$a = 1 ,b = 4 \:and \:c = 3$

$Given\:\alpha \:and \:\beta \:are \\ zeroes \:of \:p(x)$

$i ) \alpha + \beta = \frac{-b}{a} \\= \frac{ -4}{1} \\= -4 \: ---(1)$

$ii ) \alpha \times \beta = \frac{c}{a} \\= \frac{ 3}{1} \\= 3 \: ---(2)$

$If \: 1 + \frac{\beta}{\alpha } \: and \: 1 + \frac{\alpha}{\beta } \:are \: zeroes \: of \\polynomial$

$iii ) Sum \: of \:the \: Zeroes \\= 1 + \frac{\beta}{\alpha } + 1 + \frac{\alpha}{\beta }\\= 2 + \frac{\beta}{\alpha } + \frac{\alpha}{\beta }\\= \frac{2\alpha \beta + \beta^{2} + \alpha^{2}}{\alpha \beta } \\= \frac{ (\alpha+\beta)^{2}}{\alpha \beta }\\= \frac{(-4)^{2}}{3} \\= \frac{16}{3} \: --(3)$

$iv ) Product \: of \:the \: Zeroes \\=\Big( 1 + \frac{\beta}{\alpha } \Big)\Big( 1 + \frac{\alpha}{\beta }\Big)\\= 1+ \frac{\beta}{\alpha } + \frac{\alpha}{\beta }+1\\ = 2 + \frac{\beta}{\alpha } + \frac{\alpha}{\beta }\\= \frac{2\alpha \beta + \beta^{2} + \alpha^{2}}{\alpha \beta } \\= \frac{ (\alpha+\beta)^{2}}{\alpha \beta }\\= \frac{(-4)^{2}}{3} \\= \frac{16}{3} \: --(4)$

$Now , Required \: polynomial \\</p><p>k[x^{2} - \Big(1 + \frac{\beta}{\alpha } + 1 + \frac{\alpha}{\beta }\Big)x + \Big(1 + \frac{\beta}{\alpha }\Big)\Big( 1 + \frac{\alpha}{\beta }\Big)]\\= k[x^{2} - (3)x + (4)] \\= k[ x^{2} - \frac{16}{3}x + \frac{16}{3}]$

$For \:all \:real \:value \:of \:k , it \:is \:true .$

$If \: k = 3 \: then \:the \: polynomial\\ = 3x^{2} -16x + 16$

Therefore.,

$\red { Required \: polynomial} \green { = 3x^{2} -16x + 16}$

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