Question

If α and β are the zeros of the quadratic polynomial f(x) = 3x2– 4x + 5, find a polynomial whose zeros are 2α + 3β and 3α + 2β.

Answer 1

Answer:

done. have patience to solve it

Answer 2

$Given \: \alpha \:and \: \beta \: are \: zeroes$

$of \: the \: Quadratic \: polynomial$

$f(x) = 3x^{2} - 4x + 5 .$

/* Compare f(x) with ax²+bx+c , we get */

$a = 3 , b = -4 \:and \: c = 5$

$i) Sum \:of \:the \: zeroes = \frac{-b}{a}$

$\implies \alpha + \beta = \frac{-(-4)}{3}$

$= \frac{4}{3}\: ---(1)$

$ii) Product \:of \:the \: zeroes = \frac{c}{a}$

$\implies \alpha \beta = \frac{5}{3}\: --(2)$

$= \frac{4}{3}\: ---(1)$

$Given, ( 2\alpha + 3\beta) \: and \: ( 3\alpha + 2\beta)$

$are \: zeroes \: of \: another \: polynomial$

$iii) Sum \: of \: the \: zeroes$

$= ( 2\alpha + 3\beta)+( 3\alpha + 2\beta)$

$= ( 5\alpha + 5\beta)$

$= 5 ( \alpha + \beta)$

$= 5 \times \frac{4}{3}$

$= \frac{20}{3} \: --(3)$

$iv) Product\: of \: the \: zeroes$

$= ( 2\alpha + 3\beta)( 3\alpha + 2\beta)$

$= 6\alpha^{2} + 4\alpha \beta + 9\alpha \beta + 6\beta^{2}$

$= 6(\alpha^{2} + \beta)^{2} )+ 13\alpha \beta$

$= 6 [ (\alpha + \beta)^{2} - 2\alpha \beta] + 13\alpha \beta$

$= 6 (\alpha + \beta)^{2} - 12\alpha \beta+13\alpha \beta$

$= 6 (\alpha + \beta)^{2} +\alpha \beta$

$= 6 \times \Big(\frac{4}{3}\Big)^{2} + \frac{5}{3}$

$= 6 \times \frac{16}{9} + \frac{5}{3}$

$= \frac{32}{3} + \frac{5}{3}$

$= \frac{32+5}{3}$

$= \frac{37}{3}\: ---(4)$

$Now,\pink{ Required \: polynomial : }$

$\pink{= k[ x^{2} - (sum \:of \:the \: zeroes)+ product \:of \: zeroes ] }$

$= k[ x^{2} - \frac{20}{3} x + \frac{37}{3}]$

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

$If \: k= 3 \:then \: the \: polynomial \: is$

$\green { = 3x^{2} - 20x + 37 }$

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