Question

If α and β are the zeroes
p(x)= 3x^2+2x+1 find the
polynomial whose zeroes are 1-α/1+α and 1-β/1+β​

Answer 1

${\mathfrak{\red{\underline{\underline{Answer:-}}}}}$

$\sf{As\;\alpha \;and\;\beta \;are\;the\;zeroes\;of\;3x^{2}+2x+1.}$

$\sf{So,\;\alpha +\beta =\dfrac{-2}{3} \;and\;\alpha \beta =\dfrac{1}{3}}$

$\bf{Now,\;the\;sum\;of\;zeroes\;of\;the\;required\;polynomial,S=\dfrac{1-\alpha}{1+\alpha} + \dfrac{1-\beta}{1+\beta}}$

$\sf{=\dfrac{(1-\alpha)(1+\beta)+(1+\alpha)(1-\beta)}{(1+\alpha)(1+\beta)}}$

$\sf{=\dfrac{1+\beta-\alpha-\alpha \beta+1-\beta +\alpha-\alpha \beta }{1+\beta+\alpha+\alpha \beta}}$

$\sf{=\dfrac{2-2\alpha \beta}{1+\beta+\alpha+\alpha \beta}}$

$\sf{=\frac{2-2\Big(\dfrac{1}{3}\Big)}{1-\dfrac{2}{3}+\dfrac{1}{3}}}$

$\sf{=\dfrac{2-\dfrac{2}{3}}{1-\dfrac{1}{3}}}$

$\sf{=\dfrac{\Big(\dfrac{4}{3}\Big)}{\Big(\dfrac{2}{3}\Big)}}$

$\sf{\implies S =2}$

$\sf{Product\;of\;zeroes\;of\;the\;required\;polynomial,\;P=\Big(\dfrac{1-\alpha}{1+\alpha}\Big)\; \Big(\dfrac{1-\beta}{1+\beta} \Big)}$

$\sf{=\dfrac{(1-\alpha)(1-\beta)}{(1+\alpha)(1+\beta)}}$

$\sf{=\dfrac{1-\beta-\alpha+\alpha \beta }{1+\beta+\alpha+\alpha \beta}}$

$\sf{=\dfrac{1-(\beta+\alpha)+\alpha \beta }{1+(\beta+\alpha)+\alpha \beta}}$

$=\dfrac{1+\dfrac{2}{3}+\dfrac{1}{3}}{1-\dfrac{2}{3}+\dfrac{1}{3}}$

$\sf{=\dfrac{\Big(\dfrac{6}{3}\Big)}{\Big(\dfrac{2}{3}\Big)}}$

$\sf{\implies P=4}$

$\sf{Now,\;the\;required\;polynomial=x^{2}+Sx+P}$

$\sf{=x^{2}-2x+4}$

Answer 2

Solution:

Given:

➜ If α and β are the zeroes p(x) = 3x² + 2x + 1.

Find:

➜ find the Polynomial whose zeroes are 1 - α/1 + α and 1 - β1 + β.

Sum of zeroes of required polynomial :

➜ (S = 1 - α/1 + α) + (1 - β/1 + β)

➜ (1 - α) (1 + β) + (1 + α) (1 - β) / (1 + α) (1 + β) 1 + β - α - α

➜ β + 1 - β + α - αβ / 1 + β + α + αβ

➜ 2 - 2αβ / 1 + β + α + α

➜ 2 - 2 (1)/(3) / 1 - 2 / 3 + 1 / 3

➜ 2 - 2 /3 / 1 - 1/3

➜ 4/4 / 2/3

➜ S = 2

Product of zeroes of required polynomial :

➜ P = (1 - α)/ (1 + α) (1 - β)/(1 + β)

➜ (1 - α) (1 - β) / (1 + α) (1 + β)

➜ 1 - β - α + αβ / 1 + β + α + αβ

➜ 1 - (β + α) + αβ / 1 + (β + α) + αβ

➜ 1 + 2/3 + 1/3 / 1 - 2/3 + 1/3

➜ 6/3 / 2/3

➜ P = 4

Therefore, required polynomial = x² + 5x + P

=> x² - 2x + 4

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