CAT 2019

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Quadratic Equations — Vieta's Formulas (Sum and Product of Roots)
• Formula: $$ \text{For } x^2 + bx + c = 0 \text{ with roots } \alpha, \beta: \quad \alpha + \beta = -b, \quad \alpha \cdot \beta = c $$

2. Calculation

The two roots of the quadratic equation (x^2 + bx + c = 0) are given as (4a) and (3a), where (a) is an integer.

Using Vieta's formulas, the sum of roots gives us: $$4a + 3a = -b \implies 7a = -b \implies b = -7a$$

The product of roots gives us: $$4a \times 3a = c \implies c = 12a^2$$

Now we compute (b^2 + c): $$b^2 + c = (-7a)^2 + 12a^2 = 49a^2 + 12a^2 = 61a^2$$

So, (b^2 + c) must be a multiple of 61 that is also a perfect square multiple (i.e., (61a^2) for some integer (a)).

Let's verify each option by checking if dividing by 61 yields a perfect square:

$$\frac{3721}{61} = 61 \implies a^2 = 61 \implies a = \sqrt{61} \quad \text{(not an integer ❌)}$$

$$\frac{361}{61} = 5.9... \quad \text{(not an integer ❌)}$$

$$\frac{427}{61} = 7 \implies a^2 = 7 \implies a = \sqrt{7} \quad \text{(not an integer ❌)}$$

$$\frac{549}{61} = 9 \implies a^2 = 9 \implies a = 3 \quad \text{(integer ✅)}$$

With (a = 3): (b = -7(3) = -21), (c = 12(9) = 108), and (b^2 + c = 441 + 108 = 549). ✅

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3. Solution

Answer = Option 4 ✅

The final calculated value is 549, achieved when (a = 3), giving (b^2 + c = 61 \times 9 = 549).