Theory of EquationsmediumFree

Theory of Equations: Roots Equation Value

JEE Maths question with a full step-by-step solution.

Question
If α\alpha and β\beta are the roots of the equation x2−x+11=0x^2 - x + 11 = 0, then the value of 3α3−3α2+2β3−2β2+11α3\alpha^3 - 3\alpha^2 + 2\beta^3 - 2\beta^2 + 11\alpha is
A3333
B−33-33
C2222
D−22-22correct
Solution
Step 1: Since α2−α+11=0\alpha^2 - \alpha + 11 = 0, multiply by 3α3\alpha:
3α3−3α2+33α=0⋯(1)3\alpha^3 - 3\alpha^2 + 33\alpha = 0 \quad \cdots (1)
 Since β2−β+11=0\beta^2 - \beta + 11 = 0, multiply by 2β2\beta:
2β3−2β2+22β=0⋯(2)2\beta^3 - 2\beta^2 + 22\beta = 0 \quad \cdots (2)
 Step 2: Adding (1) and (2):
3α3−3α2+2β3−2β2+33α+22β=03\alpha^3 - 3\alpha^2 + 2\beta^3 - 2\beta^2 + 33\alpha + 22\beta = 0
 Step 3: Rearrange:
3α3−3α2+2β3−2β2+11α=−(22α+22β)=−22(α+β)3\alpha^3 - 3\alpha^2 + 2\beta^3 - 2\beta^2 + 11\alpha = -(22\alpha + 22\beta) = -22(\alpha + \beta)
 Step 4: By Vieta's: α+β=1\alpha + \beta = 1.
3α3−3α2+2β3−2β2+11α=−22×1=−223\alpha^3 - 3\alpha^2 + 2\beta^3 - 2\beta^2 + 11\alpha = -22 \times 1 = -22
 Correct answer: (4)
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