If the sum of the squares of the reciprocals of the roots α and β of the equation 3x² + λx - 1 = 0 i

Q: Why don't we solve for $\alpha$ and $\beta$ using the quadratic formula?

That is a classic trap! Solving for the roots directly would lead to a mess of square roots and potentially complex numbers. In JEE problems, always look for symmetry. We only need the sum and product of the roots, not the roots themselves. This approach is much cleaner and less prone to calculation errors.

Q: Why did we stop at $\lambda^2 = 9$ instead of finding $\lambda = \pm 3$?

Great observation! Notice that the final expression $6(\alpha^3 + \beta^3)^2$ involves $\lambda^2$ terms. Whether $\lambda$ is $3$ or $-3$, $\lambda^2$ remains $9$. We save time and avoid unnecessary steps by working with the square directly.

Q: Is there a faster way to handle $\alpha^3 + \beta^3$?

You could use Newton's Sums, which relate the sums of powers of roots to the coefficients of the polynomial. However, for a cubic expression, the algebraic identity $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$ is extremely efficient and less likely to cause confusion during an exam.

Question

If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$ is 15, then $6(\alpha^3 + \beta^3)^2$ is equal to:

Options:

(A) 18

(B) 24

(C) 36

(D) 96

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are going to dive into a problem that might look like a standard quadratic equation exercise, but it is actually a masterclass in algebraic symmetry.

Imagine you are standing before a parabola defined by $3x^2 + \lambda x - 1 = 0$. The roots, $\alpha$ and $\beta$, are the points where this curve kisses the x-axis.

We don't need to know exactly where they are; we only need to understand their relationship. This is the essence of **Vieta's formulas**. By using these, we bridge the gap between the roots and the coefficients:

$$\alpha + \beta = -\frac{\lambda}{3}$$

$$\alpha\beta = -\frac{1}{3}$$

These two values are the building blocks for everything that follows.

### Decoding the Condition

The problem gives us a tantalizing clue: the sum of the squares of the reciprocals of the roots is $15$. Mathematically, this is expressed as:

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$$

At first glance, this looks like a nightmare of fractions. If we find a common denominator, the expression becomes:

$$\frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = 15$$

Now, look at the numerator. We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. This is a classic identity that turns a sum of squares into a combination of the sum and product of the roots.

By substituting our Vieta's values, we transform the entire condition into a simple equation involving $\lambda^2$:

$$\frac{(-\frac{\lambda}{3})^2 - 2(-\frac{1}{3})}{(-\frac{1}{3})^2} = 15$$

After some careful arithmetic, we find that $\lambda^2 + 6 = 15$, which simplifies beautifully to $\lambda^2 = 9$. We have successfully unlocked the value of $\lambda^2$ without ever needing to find $\lambda$ itself!

### The Power of Identities

Now, we face the final challenge: calculating $6(\alpha^3 + \beta^3)^2$. We use the identity for the sum of cubes:

$$\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$$

This is a favorite concept in JEE because it tests your ability to manipulate expressions into known quantities. We substitute our sum and product values into this identity:

$$\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right) \left( \left(-\frac{\lambda}{3}\right)^2 - 3\left(-\frac{1}{3}\right) \right)$$

$$\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right) \left( \frac{\lambda^2}{9} + 1 \right)$$

Since we know $\lambda^2 = 9$, we substitute this value into the expression:

$$\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right) \left( \frac{9}{9} + 1 \right) = -\frac{2\lambda}{3}$$

Squaring this result and multiplying by $6$, we get:

$$6 \left( -\frac{2\lambda}{3} \right)^2 = 6 \left( \frac{4\lambda^2}{9} \right) = \frac{24\lambda^2}{9} = \frac{8\lambda^2}{3}$$

Substituting $\lambda^2 = 9$ into the final expression:

$$\frac{8(9)}{3} = 24$$

The final answer is **24**. This journey shows us that even the most complex-looking problems can be solved with elegance if we rely on the fundamental properties of roots.

Animated Solution for Mathematics - Quadratic Equations: If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 (α^{3} + β^{3})^{2}$ is equal to:

Select Answer:

Visualized Solution

Identify the Quadratic Equation

Apply Vieta's Formulas

Analyze the Given Condition

Simplify the Condition

Use Algebraic Identity

Substitute Vieta's Values

Solve for $λ^{2}$

Finalize $λ^{2}$

Analyze the Target Expression

Identity for Sum of Cubes

Substitute into Target

Square the Expression

Final Substitution and Compute

The Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Decoding the Condition

The Power of Identities

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