Question

If alpha, beta are the zeroes of polynomial 3x² + 2x + 1, find the polynomial whose zeroes are —

1 - alpha/1 + alpha and
1 - beta/1 + beta​

Answer 1

Answer:

The polynomial is

$x {}^{2} - \frac{x}{2}$

Refer to the attachment for the complete solution.

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Answer 2

$\huge\fcolorbox{black}{aqua}{Solution:-}$

★ Answer :-

• The required polynomial is x² - 2x + 3 .

★ Step-by-step explanation :-

• We have,

☞ α + β = - b/a = - 2/3

and

☞ α - β = c/a = 1/3

Now,

☞ The sum of the new polynomial will be,

1 - α /1 + α + 1 - β /1 + β

= 1 - α + β - αβ + 1 - β + α - αβ /(1 + α)(1 - β)

= 2 - 2αβ /1 + α + β + αβ

On plugging in the respective values, we get,

2 - 2 (1/3) /1 - 2/3 + 1/3

= 2 - 2/3 /1 - 1/3

= 4/3 /2/3

= 2

☞ Product of the new polynomial :-

[1 - α /1 + α] × [1 - β /1 + β]

= 1 - β - α + αβ /1 + β + α + αβ

= 1 - (α + β) + αβ /1 + α + β + αβ

= 1 - (- 2/3) + 1/3 /1 - 2/3 + 1/3

= 2 /2/3

= 3

∴ The required polynomial

= x² - (sum of zeroes) + product of zeroes

= x² - 2x + 3

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I hope this helps! :)