CBSE Class 10 Answered
Let α, β and γ be the roots or zeros,
of the polynomial, ax3 +3bx2 +3cx + d
then we have
α + β + γ = -3b/a
αβ+ βγ + αγ = 3c/a
αβγ = -d/a,
Since the roots are in AP,
let they be, A-D, A and A+D, where A is a term and D is common difference,
then
α + β + γ = -3b/a
A-D + A + A+D = -3b/a
3A = -3b/a
A = -b/a ... (1)
Next,
αβ+ βγ + αγ = 3c/a
(A-D)A + A(A+D) + (A-D)(A+D) = 3c/a
A2 - AD + A2 + AD + A2 - D2 = 3c/a
3A2 - D2 = 3c/a
D2 = 3A2 - c/a = 3b2/a2 - 3c/a ... from (1)
αβγ = -d/a,
(A-D)A(A+D) = -d/a
A(A2-D2) = -d/a
Put for A and D2,
(-b/a)((b2/a2 - (3b2/a2 - 3c/a)) = -d/a
(b2/a2 - (3b2/a2 - 3c/a)) = d/b
(b2 - (3b2 - 3ac)) = a2d/b
(b3 - (3b3 - 3abc)) = a2d
(-2b3 + 3abc) = a2d
2b3 - 3abc + a2d = 0
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