Skip to main content
Back to Quadratic Equations
JEE Main 2020
Quadratic Equations
Quadratic Equation and Inequalities
Hard

Question

If λ\lambda ∈\in R is such that the sum of the cubes of the roots of the equation, x 2 + (2 −- λ\lambda ) x + (10 −- λ\lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :

Options

Solution

Key Concepts and Formulas

  • Vieta's Formulas: For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 with roots α\alpha and β\beta:
    • α+β=−ba\alpha + \beta = -\frac{b}{a}
    • αβ=ca\alpha \beta = \frac{c}{a}
  • Sum of Cubes Identity: α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)
  • Difference of Roots: (α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta

Step-by-Step Solution

1. Express the Sum of Cubes in terms of λ\lambda

We are given the quadratic equation x2+(2−λ)x+(10−λ)=0x^2 + (2 - \lambda)x + (10 - \lambda) = 0. Let its roots be α\alpha and β\beta. We aim to express α3+β3\alpha^3 + \beta^3 as a function of λ\lambda.

  • Applying Vieta's Formulas: Comparing the given equation with the standard form ax2+bx+c=0ax^2 + bx + c = 0, we identify a=1a = 1, b=(2−λ)b = (2 - \lambda), and c=(10−λ)c = (10 - \lambda). Thus, α+β=−2−λ1=λ−2\alpha + \beta = -\frac{2 - \lambda}{1} = \lambda - 2 αβ=10−λ1=10−λ\alpha \beta = \frac{10 - \lambda}{1} = 10 - \lambda

  • Using the Sum of Cubes Identity: We use the identity α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) to express the sum of cubes in terms of λ\lambda: α3+β3=(λ−2)3−3(10−λ)(λ−2)\alpha^3 + \beta^3 = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2) Let f(λ)=α3+β3f(\lambda) = \alpha^3 + \beta^3. Expanding and simplifying: f(λ)=(λ3−6λ2+12λ−8)−3(10λ−20−λ2+2λ)f(\lambda) = (\lambda^3 - 6\lambda^2 + 12\lambda - 8) - 3(10\lambda - 20 - \lambda^2 + 2\lambda) f(λ)=λ3−6λ2+12λ−8−3(12λ−λ2−20)f(\lambda) = \lambda^3 - 6\lambda^2 + 12\lambda - 8 - 3(12\lambda - \lambda^2 - 20) f(λ)=λ3−6λ2+12λ−8−36λ+3λ2+60f(\lambda) = \lambda^3 - 6\lambda^2 + 12\lambda - 8 - 36\lambda + 3\lambda^2 + 60 f(λ)=λ3−3λ2−24λ+52f(\lambda) = \lambda^3 - 3\lambda^2 - 24\lambda + 52

2. Find the Minimum Value of f(λ)f(\lambda) using Calculus

To find the value of λ\lambda that minimizes f(λ)f(\lambda), we use the first and second derivative tests.

  • First Derivative Test: We calculate the first derivative of f(λ)f(\lambda) with respect to λ\lambda: f′(λ)=ddλ(λ3−3λ2−24λ+52)=3λ2−6λ−24f'(\lambda) = \frac{d}{d\lambda}(\lambda^3 - 3\lambda^2 - 24\lambda + 52) = 3\lambda^2 - 6\lambda - 24 Setting f′(λ)=0f'(\lambda) = 0 to find critical points: 3λ2−6λ−24=03\lambda^2 - 6\lambda - 24 = 0 Dividing by 3: λ2−2λ−8=0\lambda^2 - 2\lambda - 8 = 0 Factoring the quadratic equation: (λ−4)(λ+2)=0(\lambda - 4)(\lambda + 2) = 0 The critical points are λ=4\lambda = 4 and λ=−2\lambda = -2.

  • Second Derivative Test: The second derivative of f(λ)f(\lambda) with respect to λ\lambda is: f′′(λ)=ddλ(3λ2−6λ−24)=6λ−6f''(\lambda) = \frac{d}{d\lambda}(3\lambda^2 - 6\lambda - 24) = 6\lambda - 6 Evaluating f′′(λ)f''(\lambda) at the critical points:

    • For λ=−2\lambda = -2: f′′(−2)=6(−2)−6=−12−6=−18<0f''(-2) = 6(-2) - 6 = -12 - 6 = -18 < 0 Thus, f(λ)f(\lambda) has a local maximum at λ=−2\lambda = -2.
    • For λ=4\lambda = 4: f′′(4)=6(4)−6=24−6=18>0f''(4) = 6(4) - 6 = 24 - 6 = 18 > 0 Thus, f(λ)f(\lambda) has a local minimum at λ=4\lambda = 4.

Therefore, the sum of the cubes of the roots is minimized when λ=4\lambda = 4.

3. Formulate the Quadratic Equation for the minimum λ\lambda

Substituting λ=4\lambda = 4 into the original quadratic equation: x2+(2−4)x+(10−4)=0x^2 + (2 - 4)x + (10 - 4) = 0 x2−2x+6=0x^2 - 2x + 6 = 0

4. Calculate the Magnitude of the Difference of Roots

For the equation x2−2x+6=0x^2 - 2x + 6 = 0:

  • Applying Vieta's Formulas: α+β=−−21=2\alpha + \beta = -\frac{-2}{1} = 2 αβ=61=6\alpha \beta = \frac{6}{1} = 6

  • Using the Difference of Roots Formula: (α−β)2=(α+β)2−4αβ=(2)2−4(6)=4−24=−20(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (2)^2 - 4(6) = 4 - 24 = -20 (α−β)=±−20=±25i(\alpha - \beta) = \pm \sqrt{-20} = \pm 2\sqrt{5}i

  • Finding the Magnitude: The magnitude of the difference of the roots is ∣α−β∣=∣±25i∣=02+(25)2=20=25|\alpha - \beta| = | \pm 2\sqrt{5}i | = \sqrt{0^2 + (2\sqrt{5})^2} = \sqrt{20} = 2\sqrt{5}

Common Mistakes & Tips

  • Be careful with signs when applying Vieta's formulas.
  • Double-check the differentiation steps in the calculus part.
  • Remember that a negative discriminant implies complex roots.

Summary

We found the sum and product of the roots in terms of λ\lambda using Vieta's formulas. We then expressed the sum of cubes of the roots as a function of λ\lambda, f(λ)=λ3−3λ2−24λ+52f(\lambda) = \lambda^3 - 3\lambda^2 - 24\lambda + 52. Using calculus, we minimized f(λ)f(\lambda) and found that λ=4\lambda = 4. Substituting this value into the original equation gave x2−2x+6=0x^2 - 2x + 6 = 0. Finally, we found the magnitude of the difference of the roots to be 252\sqrt{5}.

The final answer is 25\boxed{2\sqrt{5}}, which corresponds to option (B).

Previous QuestionNext Question

Practice More Quadratic Equations Questions

View All Questions