Theory of EquationsmediumFree

Theory of Equations: Equation Roots Value Equal

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

Question
If the equation x3+2x2−4x+5=0x^3 + 2x^2 - 4x + 5 = 0 has roots α,β,γ\alpha, \beta, \gamma, then the value of (α3+5)(β3+5)(γ3+5)13αβγ\dfrac{(\alpha^3+5)(\beta^3+5)(\gamma^3+5)}{13\alpha\beta\gamma} is equal to
A55
B88correct
C1212
D1515
Solution
Step 1: From x3+2x2−4x+5=0x^3 + 2x^2 - 4x + 5 = 0: each root satisfies α3=−2α2+4α−5\alpha^3 = -2\alpha^2 + 4\alpha - 5.
α3+5=−2α2+4α=2α(2−α)\alpha^3 + 5 = -2\alpha^2 + 4\alpha = 2\alpha(2 - \alpha)
 Similarly: β3+5=2β(2−β)\beta^3 + 5 = 2\beta(2-\beta) and γ3+5=2γ(2−γ)\gamma^3 + 5 = 2\gamma(2-\gamma). Step 2: By Vieta's: αβγ=−5\alpha\beta\gamma = -5 (constant term with sign: −5/1-5/1). Step 3: Evaluate (2−α)(2−β)(2−γ)(2-\alpha)(2-\beta)(2-\gamma) by substituting x=2x = 2 into the polynomial:
23+2(4)−4(2)+5=8+8−8+5=132^3 + 2(4) - 4(2) + 5 = 8 + 8 - 8 + 5 = 13
 Step 4:
(α3+5)(β3+5)(γ3+5)=8αβγ(2−α)(2−β)(2−γ)=8⋅(−5)⋅13=−520(\alpha^3+5)(\beta^3+5)(\gamma^3+5) = 8\alpha\beta\gamma(2-\alpha)(2-\beta)(2-\gamma) = 8 \cdot (-5) \cdot 13 = -520
−52013αβγ=−52013⋅(−5)=−520−65=8\frac{-520}{13\alpha\beta\gamma} = \frac{-520}{13 \cdot (-5)} = \frac{-520}{-65} = 8
 Correct answer: (2)
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