Skip to main content
Back to Quadratic Equations
JEE Main 2024
Quadratic Equations
Quadratic Equation and Inequalities
Hard

Question

Let α,β,γ\alpha, \beta, \gamma be the three roots of the equation x3+bx+c=0x^{3}+b x+c=0. If βγ=1=α\beta \gamma=1=-\alpha, then b3+2c33α36β38γ3b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3} is equal to :

Options

Solution

Key Concepts and Formulas

  • Vieta's Formulas: For a cubic equation x3+bx+c=0x^3 + bx + c = 0 with roots α,β,γ\alpha, \beta, \gamma:
    • α+β+γ=0\alpha + \beta + \gamma = 0
    • αβ+βγ+γα=b\alpha\beta + \beta\gamma + \gamma\alpha = b
    • αβγ=c\alpha\beta\gamma = -c
  • Roots of Unity: The solutions to x3=1x^3 = -1 are 1,ω,ω2-1, -\omega, -\omega^2, where ω=ei2π/3\omega = e^{i2\pi/3} is a complex cube root of unity. Key properties include ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0.

Step-by-Step Solution

Step 1: Determine the value of cc using the product of roots.

We are given that α,β,γ\alpha, \beta, \gamma are the roots of x3+bx+c=0x^3 + bx + c = 0. From Vieta's formulas, the product of the roots is αβγ=c\alpha\beta\gamma = -c. We are also given that βγ=1\beta\gamma = 1 and α=1\alpha = -1.

  • Reasoning: The product of the roots formula relates the roots to the constant term cc. We can use the given conditions for α,β,γ\alpha, \beta, \gamma to find the value of cc.

Substituting α=1\alpha = -1 and βγ=1\beta\gamma = 1 into the product of roots formula: (1)(1)=c(-1)(1) = -c 1=c-1 = -c Therefore, c=1c = 1

Step 2: Determine the value of bb using the fact that α\alpha is a root.

Since α=1\alpha = -1 is a root of the equation x3+bx+c=0x^3 + bx + c = 0, it must satisfy the equation when substituted for xx.

  • Reasoning: If a value is a root of an equation, substituting it into the equation will result in the equation being equal to zero. This allows us to find the coefficient bb.

Substituting x=1x = -1 into the equation: (1)3+b(1)+c=0(-1)^3 + b(-1) + c = 0 1b+c=0-1 - b + c = 0 Substituting the value of c=1c = 1 that we found in Step 1: 1b+1=0-1 - b + 1 = 0 b=0-b = 0 Therefore, b=0b = 0

Step 3: Identify the cubic equation and its roots.

Now that we have b=0b = 0 and c=1c = 1, the original cubic equation x3+bx+c=0x^3 + bx + c = 0 becomes: x3+(0)x+1=0x^3 + (0)x + 1 = 0 x3+1=0x^3 + 1 = 0 This can be rewritten as x3=1x^3 = -1.

  • Reasoning: By finding the coefficients bb and cc, we have determined the specific cubic equation. Recognizing this equation as x3=1x^3 = -1 is crucial because its roots are standard complex numbers related to the cube roots of unity.

The roots of x3=1x^3 = -1 are 1,ω,ω2-1, -\omega, -\omega^2, where ω\omega is a complex cube root of unity. We already know that one root is α=1\alpha = -1. Therefore, the other two roots, β\beta and γ\gamma, must be ω-\omega and ω2-\omega^2. Let's verify this assignment with the given condition βγ=1\beta\gamma = 1: If we assign β=ω\beta = -\omega and γ=ω2\gamma = -\omega^2: βγ=(ω)(ω2)=ω3\beta\gamma = (-\omega)(-\omega^2) = \omega^3 Since ω3=1\omega^3 = 1, we have: βγ=1\beta\gamma = 1 This is consistent with the given condition. So, we have:

  • α=1\alpha = -1
  • β=ω\beta = -\omega
  • γ=ω2\gamma = -\omega^2

Step 4: Evaluate the given expression.

We need to calculate the value of the expression: b3+2c33α36β38γ3b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3

  • Reasoning: With all the roots and coefficients determined, we can now substitute these values into the target expression. The properties of ω\omega will be essential for simplifying terms like (ω)3(-\omega)^3 and (ω2)3(-\omega^2)^3.

Substituting the values: b=0b = 0, c=1c = 1, α=1\alpha = -1, β=ω\beta = -\omega, γ=ω2\gamma = -\omega^2: (0)3+2(1)33(1)36(ω)38(ω2)3(0)^3 + 2(1)^3 - 3(-1)^3 - 6(-\omega)^3 - 8(-\omega^2)^3 Simplifying each term:

  • 03=00^3 = 0
  • 2(1)3=2(1)=22(1)^3 = 2(1) = 2
  • 3(1)3=3(1)=33(-1)^3 = 3(-1) = -3
  • 6(ω)3=6(1)3(ω3)=6(1)(1)=66(-\omega)^3 = 6(-1)^3(\omega^3) = 6(-1)(1) = -6
  • 8(ω2)3=8(1)3(ω6)=8(1)(ω3)2=8(1)(1)2=88(-\omega^2)^3 = 8(-1)^3(\omega^6) = 8(-1)(\omega^3)^2 = 8(-1)(1)^2 = -8

Now, substitute these simplified terms back into the expression: 0+2(3)(6)(8)0 + 2 - (-3) - (-6) - (-8) 0+2+3+6+80 + 2 + 3 + 6 + 8 1919

Common Mistakes & Tips

  • Sign Errors in Vieta's Formulas: Ensure correct signs. The product of roots is D/A-D/A for Ax3+Bx2+Cx+D=0Ax^3+Bx^2+Cx+D=0.
  • Incorrect Identification of Cube Roots of Unity: Remember the roots of x3=1x^3 = -1 are 1,ω,ω2-1, -\omega, -\omega^2.
  • Powers of ω\omega: Use ω3=1\omega^3 = 1 to simplify higher powers.

Summary

We used Vieta's formulas and the properties of cube roots of unity to solve this problem. We first determined the coefficients bb and cc of the cubic equation. Then, using the fact that x3=1x^3 = -1, we identified the roots α,β,γ\alpha, \beta, \gamma. Finally, we substituted these values into the given expression and simplified to find the result.

The final answer is \boxed{19}, which corresponds to option (B).

Practice More Quadratic Equations Questions

View All Questions