CAT 2024 Slot 2 QA Question 15

The question is about a quadratic equation: 3x² + λx - 1 = 0. The roots of this equation are α and β. We're given that 1/α² + 1/β² equals 15, and we need to find the value of (α³ + β³)². The options are A) 16, B) 4, C) 1, D) 9.

Alright, so first, for a quadratic equation ax² + bx + c = 0, the sum of the roots α + β is equal to -b/a, and the product αβ is c/a. That's from Vieta's formulas. So in this case, the equation is 3x² + λx - 1 = 0, so a = 3, b = λ, c = -1.

So, let's write down what we know:

Sum of roots, α + β = -λ/3.

Product of roots, αβ = -1/3.

Now, the given condition is 1/α² + 1/β² = 15. Hmm, I need to relate this to the sum and product of the roots. 1/α² + 1/β² can be expressed in terms of (α + β) and αβ. Let me think about how.

I recall that 1/α² + 1/β² is equal to (α² + β²)/(αβ)². Wait, is that right? Let me check:

1/α² + 1/β² = (β² + α²)/(αβ)². Yes, that's correct. So, that's equal to (α² + β²)/(αβ)².

But I also know that α² + β² can be written in terms of (α + β)² - 2αβ. So, substituting that in, we have:

1/α² + 1/β² = [(α + β)² - 2αβ]/(αβ)².

So, plugging in the values we have:

[(α + β)² - 2αβ]/(αβ)² = 15.

We already know α + β = -λ/3 and αβ = -1/3. Let's substitute those in.

First, compute (α + β)²:

(-λ/3)² = λ²/9.

Then, compute 2αβ:

2*(-1/3) = -2/3.

So, the numerator becomes:

λ²/9 - (-2/3) = λ²/9 + 2/3.

The denominator is (αβ)² = (-1/3)² = 1/9.

So, putting it all together:

(λ²/9 + 2/3) / (1/9) = 15.

Simplify the division:

(λ²/9 + 2/3) * 9 = 15.

Multiplying each term by 9:

λ² + 6 = 15.

Subtract 6 from both sides:

λ² = 9.

So, λ can be either 3 or -3. Hmm, okay, so λ is either 3 or -3.

Now, the question asks for (α³ + β³)². We find this value. Let me recall the formula for α³ + β³.

α³ + β³ = (α + β)(α² - αβ + β²). Alternatively, it can also be written as (α + β)^3 - 3αβ(α + β). Let me verify which one is more useful here.

Given that I already have α + β and αβ, maybe the second formula is better because it directly uses those terms.

So, α³ + β³ = (α + β)^3 - 3αβ(α + β).

Compute each part step by step.

First, compute (α + β)^3:

(α + β)^3 = (-λ/3)^3 = (-λ)^3 / 27 = -λ³ / 27.

Then, compute 3αβ(α + β):

3αβ(α + β) = 3(-1/3)(-λ/3) = 3*(λ/9) = λ/3.

So, putting it all together:

α³ + β³ = (-λ³ / 27) - (λ/3).

Wait, hold on, The formula is (α + β)^3 - 3αβ(α + β). So, substituting:

= (-λ/3)^3 - 3(-1/3)(-λ/3)

= (-λ³ / 27) - 3(1/3)(λ/3)

Wait, Compute 3αβ(α + β):

3αβ(α + β) = 3(-1/3)(-λ/3) = 3*(λ/9) = λ/3.

So, the formula becomes:

α³ + β³ = (-λ³ / 27) - (λ/3).

Wait, that seems a bit complicated. Maybe I should use the other formula: α³ + β³ = (α + β)(α² - αβ + β²).

But then I need α² + β², which I can compute from (α + β)^2 - 2αβ.

Wait, Computing α² + β² first:

α² + β² = (α + β)^2 - 2αβ = (λ²/9) - 2*(-1/3) = λ²/9 + 2/3.

So, α² + β² = λ²/9 + 2/3.

Then, α² - αβ + β² = (α² + β²) - αβ = (λ²/9 + 2/3) - (-1/3) = λ²/9 + 2/3 + 1/3 = λ²/9 + 1.

So, α³ + β³ = (α + β)(α² - αβ + β²) = (-λ/3)(λ²/9 + 1).

Compute that:

= (-λ/3)(λ²/9 + 1) = (-λ/3)(λ² + 9)/9 = (-λ(λ² + 9))/27.

So, α³ + β³ = (-λ(λ² + 9))/27.

Now, we need to compute (α³ + β³)^2.

So, squaring both sides:

(α³ + β³)^2 = [(-λ(λ² + 9))/27]^2 =