Thread: Roots of a Cubic Polynomial in Arithmetic Sequence

Been having some difficulty with the following question:

Prove that if the roots of:

x^3-ax^2+bx-c=0 are in arithmetic sequence then 2a^3-9ab+27c=0

Hence, solve for x in the equation x^3 -12x^2+39x -28=0

I've tried putting general forms of the roots as:

(x-a),(x-(a+d)),(x-(a+2d))

and then substituting them into the initial equation and setting = 0, but it hasn't gotten me anywhere.

Any help would be greatly appreciated

Denote the roots x_1 = u-h, x_2 = u , x_3 = u+h, apply the Cardano-Vieta relations and elimitate u and h.

Originally Posted by Wandering

Been having some difficulty with the following question:

Prove that if the roots of:

x^3-ax^2+bx-c=0 are in arithmetic sequence then 2a^3-9ab+27c=0

Hence, solve for x in the equation x^3 -12x^2+39x -28=0

I've tried putting general forms of the roots as:

(x-a),(x-(a+d)),(x-(a+2d))

and then substituting them into the initial equation and setting = 0, but it hasn't gotten me anywhere.

Any help would be greatly appreciated

Alternatively,

let the roots be E, F, G in arithmetic sequence.

Then

(x-E)(x-F)(x-G) expands out to x^3-(E+F+G)x^2+(EF+EG+FG)x-EFG

which is x^3-ax^2+bx-c

Since the roots are in arithmetic sequence, then E+F+G=3F

so a=3F and b=EF+EG+FG and c=EFG

This allows 2a^3-9ab+27c to simplify to 2(3F)^3-9(3F)(EF+EG+FG)+27c

giving 54F^3-27F^2(E)-27EFG-27F^2(G)+27EFG

=54F^3-27F^2(E+G)

and since E+G=2F, the result follows.

(sorry, this would be neater with Latex).