Question

✠ If the zeros of polynomial x^3 - 3x^2 + x + 1 are a - b, a, a + b, find a and b ?

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Answer 1

Answer:

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Answer 2

Given :

$\sf \: P(x)=x^3-3x^2+x+1$

$\implies \sf \: Zeroes \: are \: \: a - b, a + b$

$\small\fbox\red{Comparing the given polynomial with PX2+QX2+RX+1, we obtain}$

$\sf P = 1, Q = -3, R = 1, T = 1$

$\small\fbox\green{Sum of zeroes = a - b + a + a + b}$

$\implies \sf \frac{-q \: }{p} = 3a$

$\implies\sf \frac{ - ( - 3)}{1} = 3a$

$\implies \sf3 = 3a$

$\implies\sf a = 1$

$\small\fbox\pink{The \: zeroes \: are 1-b, 1, 1+b}$

✠ Multiplication of zeroes = 1 (1 - b) (1 + b)

$\implies \bf\frac{ - t \: }{p} = 1- {b}^{2}$

$\implies\sf \frac{ - 1 \: }{1} =1- {b}^{2}$

$\implies \sf1 - {b}^{2} =-1$

$\implies \sf1 + 1 = {b}^{2}$

$\implies \sf \: b = ± \sqrt{2}$

$\boxed{Hence, a = 1 \: and \: b = \sqrt{2} \: or \: \sqrt{ - 2} }$