Math, asked by bibekkumar9775, 1 year ago

If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

Answers

Answered by max20
14
Let p,q,r be the roots of the cubic eqn.
As they are in AP => 2q=p+ r
=> 3q= p+q+r

p+q+r= sum of roots = a
So 3q= a => q=a/3
Let f(x)= x^3-ax^2+bx-c
So as q is a root of f(x) , f(q)=0
or, f(a/3)=0
So put a/3 in given cubic eqn and equate it to 0.
On solving we get 2a^3-9ab+27c=0
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