Question

find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial :

ax2+bx+c a and c are not equal to 0

Answer 1

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Answer 2

Answer:

Required polynomial is cx²+bx+a

Step-by-step explanation:

$Let\:\alpha \:and\:\beta\:are\\zeroes \:of \: quadratic\: equation\\ax^{2}+bx+c, \:a≠0, c≠0$

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

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

$iii)\frac{1}{\alpha}+\frac{1}{\beta}$

$=\frac{\beta+\alpha}{\alpha \beta}$

$=\frac{\frac{-b}{a}}{\frac{c}{a}}\\=\frac{-b}{c}\:--(3)$

$iv) \frac{1}{\alpha}\times \frac{1}{\beta}\\ = \frac{1}{\alpha \beta}\\=\frac{1}{\frac{c}{a}}\\=\frac{a}{c}\:--(4)$

$Now,\\Form \: of \: a\: polynomial \:whose \\zeroes \: are \: \frac{1}{\alpha}\:and\:\frac{1}{\beta}\\(Given \: reciprocals \:of \: above \: polynomial )\\is:$

$k[x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\alpha \beta ]$

$=k[x^2-\left(\frac{\beta+\alpha}{\alpha \beta}\right)+\alpha \beta]$

$=k[x^{2}-\left(\frac{-b}{c}\right)x+\frac{a}{c}]$

/* From (3) and (4) */

$=k[x^{2}+\frac{b}{c}x +\frac{a}{c}]$

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

If k = c , we get

Required polynomial is cx²+bx+a

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