Question

Find the condition that the zeroes of the polynomial f(x) = x3 - px2 + qx -r may be in ap

Answer 1

let zeroes be a-d, a, a+d

sum of zeroes = p

So a- d + a + a + d = p

3a = p

a = p/3

product of zeroes = r

( a-d)( a)( a+d) = r

a ( a^2 - d^2) = r

p/3 ( p^2/9 - d^2) = r

p ^2 /9 - d^2 = 3r/p

d^2 = p^2/9 - 3r/p


Sum of product of roots taking 2 at a time = q

a( a- D) + a( a+ d) + ( a-d)( a+ d) = q

a( a-d + a + d) + a^2 - d^2 = q

3a + a^2 - d^2 = q

3( p/3) + p^2/9 - p^2/9 + 3r/p = q

p + 3r/p = q

p^2 + 3r - pq = 0



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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

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f(x) = x³ - px² +  qx - r

Let us assumed that a-d , a and a+d be the zeroes of the polynomial f(x).

Sum of the zeroes ( $\alpha+\beta$ =  -b/a

( a - d ) + a + ( a + d ) = -(-p)/1

3a = p

a = p/3.

"a" is a zero of the polynomial f(x).

f(a) = 0

a³ - pa² + qa - r = 0

( p/3 )³ - p (p/3 )² + q ( p/3 ) - r = 0

p³ - 3p³ + 9pq - 27r = 0

2p³ - 9pq + 27r = 0.

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