Question

IF α,β ARE THE TWO ZEROES OF THE POLYNOMIAL f(x)=1x²+mx+n,then find a quadratic polynomial whose zeroes are α/β&β/a.

SPECIAL THANKS TO EVERYONE AND MSYTICD...

faa4bc25b97663004a0bf66735362799.pdf

Answer 1

$Compare \: given \: polynomial \:1x^{2}+mx+n\\with \: ax^{2}+bx+c , we \:get$

$a = 1 , b = m \:and \:c = n$

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

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

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

$Now, If \:\frac{\alpha}{\beta} \:and \: \frac{\beta}{\alpha}\:are \: Zeroes \: Quadratic \: polynomial$

$Sum \: of \:the \: zeroes\\ = \frac{\alpha}{\beta}+\frac{\beta}{\alpha}\\= \frac{\alpha^{2} + \beta^{2}}{\alpha \beta} \\= \frac{ ( \alpha + \beta )^{2} - 2\alpha \beta }{\alpha \beta } \\= \frac{ (-m)^{2} - 2n}{n} \\= \frac{m^{2}-2n}{n} \: --(3)$

$Product \:of \:the \: zeroes \\= \frac{\alpha}{\beta}\times \frac{\beta}{\alpha}\\= 1$

$Required \: polynomial$

$k[x^{2} - \Big(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\Big)x + \frac{\alpha}{\beta}\times \frac{\beta}{\alpha}]\\= k[x^{2} - \frac{m^{2}-2n}{n} + 1 ]$

/* For all real values of k it is true */

$If \: k = n \: then \\nx^{2} - (m^{2}-2n)x + n$

•••♪

Answer 2

Hope it helps...

Plz mark as brainliest....