Answer for the 9th question 9 α, β, γ are zeroes of cubic polynomial p(x) = x3 - Maths - Polynomials

Question

Answer for the 9th question

9 α ,   β ,   γ are zeroes of cubic
polynomial p(x) = x3 −
ax 2 + bx − c, such
that 2 β = α + γ Show that 2 a 3 - 9 a b + 27 c =
0

Priyanka Sahu · Accepted Answer

Dear Student,

$l e t t h e g i v e n p o l y n o m i a l b e f (x) = x^{3} - a x^{2} + b x - c$
It is mentioned that 2Beta = Alpha + Gamma
This implies that the zeroes are in A.P.

let A-d, A and A+d be the zeros of the polynomial f(x) . then

$s u m o f z e r o s = \frac{c o e f f . o f x^{2}}{co e f f . o f x^{3}} = \frac{- (- a)}{1} A - d + A + A + d = a 3 A = a A = \frac{a}{3} .... (i)$

$\sin c e A i s a z e r o o f p o l y n o m i a l f (x) t h e r e f o r e, f (x) = x^{3} - a x^{2} + b x - c f (A) = 0 f (A) = {(\frac{a}{3})}^{3} - a {(\frac{a}{3})}^{2} + b (\frac{a}{3}) - c 0 = \frac{a^{3}}{27} - \frac{a^{3}}{9} + \frac{a b}{3} - c 0 = \frac{a^{3} - 3 a^{3} + 9 a b - 27 c}{27} 0 = - 2 a^{3} + 9 a b - 27 c o r 2 a^{3} - 9 a b + 27 c = 0 h e n c e p r o v e d .$
Regards

Answer for the 9th question. 9. α , β , γ are zeroes of cubic polynomial p(x) = x3 − ax 2 + bx − c, such that 2 β = α + γ . Show that 2 a 3 - 9 a b + 27 c = 0 .

Answer for the 9th question.

9. $α, β, γ$ are zeroes of cubic polynomial p(x) = x³ − ax ² + bx − c, such that $2 β = α + γ .$ Show that $2 a^{3} - 9 a b + 27 c = 0 .$