If λ ∈ R is such that the sum of the cubes of the roots of the equation, x²+(2–λ)x+(10–λ) = 0 is min

Q: Wait, why can't I just use the quadratic formula to find $x$ and then cube them? It seems more direct.

While that sounds intuitive, it's a trap! The quadratic formula will give you expressions involving square roots of $\lambda$. Cubing those expressions will lead to an algebraic nightmare. Vieta's relations allow us to bypass the roots entirely and work directly with the coefficients, which is the elegant, 'JEE-style' way to solve this.

Q: I got $(\alpha - \beta)^2 = -20$. Does this mean I made a mistake? Can the difference of roots be imaginary?

You haven't made a mistake! In JEE Advanced, you must be comfortable with complex numbers. If the discriminant is negative, the roots are complex conjugates. The magnitude of the difference $|\alpha - \beta|$ is perfectly well-defined even when the roots themselves are not real. Trust your algebra.

Q: Why do we need the second derivative test? Can't we just pick the smaller value of $\lambda$?

Never assume! The critical points are just candidates. The function could have a local maximum at one and a local minimum at the other. The second derivative test is your mathematical compass—it tells you exactly where the 'valley' of the function lies.

Question

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

Options:

(A) $4 \sqrt{2}$

(B) 20

(C) $2 \sqrt{7}$

(D) $2 \sqrt{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given quadratic equation is $x^2 + (2 - \lambda)x + (10 - \lambda) = 0$. We are tasked with minimizing the sum of the cubes of the roots, $S = \alpha^3 + \beta^3$.

Using **Vieta's relations**, we identify the sum and product of the roots:
$$\alpha + \beta = \lambda - 2$$
$$\alpha\beta = 10 - \lambda$$

### The Master Equation

To express $S$ in terms of $\lambda$, we utilize the algebraic identity:
$$S = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

Substituting the Vieta expressions into this identity, we obtain:
$$S(\lambda) = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2)$$

Expanding this expression, we get:
$$S(\lambda) = (\lambda^3 - 6\lambda^2 + 12\lambda - 8) - 3(10\lambda - 20 - \lambda^2 + 2\lambda)$$
$$S(\lambda) = \lambda^3 - 3\lambda^2 - 24\lambda + 52$$

### Optimization via Calculus

To find the minimum value, we differentiate $S(\lambda)$ with respect to $\lambda$:
$$\frac{dS}{d\lambda} = 3\lambda^2 - 6\lambda - 24$$

Setting the derivative to zero to find the critical points:
$$3(\lambda^2 - 2\lambda - 8) = 0$$
$$3(\lambda - 4)(\lambda + 2) = 0$$

The critical points are $\lambda = 4$ and $\lambda = -2$. We apply the **second derivative test** to determine the nature of these points:
$$\frac{d^2S}{d\lambda^2} = 6\lambda - 6$$

For $\lambda = 4$, the second derivative is $6(4) - 6 = 18 > 0$, confirming a **local minimum**. For $\lambda = -2$, the second derivative is $6(-2) - 6 = -18 < 0$, confirming a local maximum.

### Final Calculation

With $\lambda = 4$, we find the sum and product of the roots:
$$\alpha + \beta = 4 - 2 = 2$$
$$\alpha\beta = 10 - 4 = 6$$

We use the identity for the square of the difference of the roots:
$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$
$$(\alpha - \beta)^2 = (2)^2 - 4(6) = 4 - 24 = -20$$

The difference is $\alpha - \beta = \sqrt{-20} = 2i\sqrt{5}$. The magnitude of the difference is:
$$|\alpha - \beta| = |2i\sqrt{5}| = \mathbf{2\sqrt{5}}$$

Animated Solution for Mathematics - Quadratic Equations: If $λ \in R$ is such that the sum of the cubes of the roots of the equation, $x^{2} + (2 - λ) x + (10 - λ) = 0$ is minimum, then the magnitude of the difference of the roots of this equation is :

Select Answer:

Visualized Solution

Defining the Equation and Roots

Applying Vieta's Relations

Expanding the Sum of Cubes

Substituting $λ$ Expressions

Simplifying the Function $S (λ)$

Finding Critical Points

Solving for $λ$

Second Derivative Test

Calculating Roots at $λ = 4$

Finding the Difference of Roots

Final Magnitude Calculation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Master Equation

Optimization via Calculus

Final Calculation

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