Question

if alpha and beta are zeroes of polynomial f(x)=x^2-1,find a quadratic polynomial whose zeroes are 2alpha/beta and 2beta/alpha

Answer 1

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Answer 2

Answer:

$The \: quadratic \: polynomial \: will\\ \: be \: x^{2}+4x+4$

Step-by-step explanation:

$Given, \\\alpha \: and \: \beta \\\: are \: zeroes \: of \: \\the \: polynomial \: f(x)=x^{2}-1$

Compare f(x)= x²-1 with ax²+bx+c ,we get

a = 1 , b = 0 , c = -1

$i)\alpha+\beta\\= \frac{-b}{a}\\=\frac{0}{1}\\=0--(1)$

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

$iii)\frac{2\alpha}{\beta}+\frac{2\beta}{\alpha}\\=\frac{2(\alpha^{2}+\beta^{2})}{\alpha\beta}\\=\frac{2[(\alpha+\beta)^{2}-2\alpha\beta]}{\alpha\beta}\\=\frac{2[0-2(-1)]}{(-1)}\\=\frac{4}{(-1)}\\=-4 --(3)$

$iv)(\frac{2\alpha}{\beta})(\frac{2\beta}{\alpha})\\=4--(4)$

Therefore,

The quadratic polynomial is

$k[x^{2}-\big(\frac{2\alpha}{\beta}+\frac{2\beta}{\alpha}\big)+)(\frac{2\alpha}{\beta})(\frac{2\beta}{\alpha})]\\, \: where \: k \: is \: a \: constant$

$=k[x^{2}-(-4)x+4$

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

$We\:can\: put \: different \: values\: of \: k$

$When \: k = 1 ,\: \\the \: quadratic \: polynomial \: will\\ \: be \: x^{2}+4x+4$

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