Theory of EquationsmediumFree

Theory of Equations: Equation Roots Value Equal

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

Question

If the equation

x^3 + 2x^2 - 4x + 5 = 0

has roots

\alpha, \beta, \gamma

, then the value of

\dfrac{(\alpha^3+5)(\beta^3+5)(\gamma^3+5)}{13\alpha\beta\gamma}

is equal to

5

8

correct

12

15

Solution

Step 1: From

x^3 + 2x^2 - 4x + 5 = 0

: each root satisfies

\alpha^3 = -2\alpha^2 + 4\alpha - 5

\alpha^3 + 5 = -2\alpha^2 + 4\alpha = 2\alpha(2 - \alpha)

Similarly:

\beta^3 + 5 = 2\beta(2-\beta)

and

\gamma^3 + 5 = 2\gamma(2-\gamma)

. Step 2: By Vieta's:

\alpha\beta\gamma = -5

(constant term with sign:

-5/1

). Step 3: Evaluate

(2-\alpha)(2-\beta)(2-\gamma)

by substituting

x = 2

into the polynomial:

2^3 + 2(4) - 4(2) + 5 = 8 + 8 - 8 + 5 = 13

Step 4:

(\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

\frac{-520}{13\alpha\beta\gamma} = \frac{-520}{13 \cdot (-5)} = \frac{-520}{-65} = 8

Correct answer: (2)

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