Prove that if the roots of x^3 - 9x^2 + 11x + c form an arithmetic sequence, then c = 21.
+129847 Let the roots = x, y, z in that order
And let c = 21
By Vieta
x + y + z = 9 ⇒ y + z = 9 - x (1)
xy + xz + yz = 11 ⇒ x ( y + z) + yz = 11 ⇒ x ( 9 - x) + yz = 11 (2)
xyz = -21 ⇒ yz = -21 / x (3)
Sub (3) into (2)
x (9 - x) - 21/x = 11
-x^2 + 9x - 21/x = 11 multiply through by -x
x^3 - 9x^2 + 21 = -11x
x^3 - 9x^2 + 11x + 21 = 0
By inspection x = -1 is a root
So
y + z = 9 - - 1 = 10
And
yz = -21/ -1 = 21
To form an arithmetic series, y = 3 and z = 7
So.....the series is
x y z
-1 3 7
