CAT 2021

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Quadratic Equations — Irrational Conjugate Root Theorem & Vieta's Formulas
• Formula: $$\text{If } a, b, c \in \mathbb{Q} \text{ and one root is } p + \sqrt{q}, \text{ then the other root must be } p - \sqrt{q}$$ $$\text{Sum of roots} = \frac{b}{a}, \quad \text{Product of roots} = \frac{c}{a}$$

---

2. Calculation

Since (a, b, c) are all rational numbers, and one root is (2 + \sqrt{3}) (an irrational number), the Irrational Conjugate Root Theorem guarantees that the other root must be (2 - \sqrt{3}). This ensures both the sum and product of roots remain rational.

Step 1: Find the sum of the roots.

$$\text{Sum} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$$

For the equation (ax^2 - bx + c = 0), by Vieta's formulas:

$$\text{Sum of roots} = \frac{b}{a} = 4 \implies b = 4a$$

Step 2: Find the product of the roots.

$$\text{Product} = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$$

By Vieta's formulas:

$$\text{Product of roots} = \frac{c}{a} = 1 \implies c = a$$

Step 3: Apply the condition (b = c^3).

Substituting (b = 4a) and (c = a):

$$4a = a^3$$

$$a^3 - 4a = 0$$

$$a(a^2 - 4) = 0$$

Since (a eq 0):

$$a^2 = 4 \implies a = 2 \text{ or } a = -2$$

Step 4: Find (|a|).

$$|a| = 2$$

---

3. Solution

Answer = Option 2 ✅

The final calculated value of (|a|) is 2.