Let α, β be the roots of the quadratic equation 12x² - 20x + 3λ = 0, λ ∈ Z. If 1/2 ≤ |β - α| ≤ 3/2,

Q: Wait, why is the formula for the difference of roots $|\beta - \alpha| = \frac{\sqrt{D}}{|a|}$? Can't I just find the roots using the quadratic formula and subtract them?

You absolutely could! But in competitive exams like JEE, time is your most precious resource. The quadratic formula involves calculating $\frac{-b \pm \sqrt{D}}{2a}$, which is prone to arithmetic errors. Using the identity $|\beta - \alpha| = \frac{\sqrt{D}}{|a|}$ is a powerful, elegant shortcut that bypasses the need to find the individual roots entirely. It is a hallmark of a seasoned problem solver.

Q: Why do we have to flip the inequality sign when dividing by $-9$?

This is a fundamental rule of inequalities that often trips up even the best students. When you multiply or divide an inequality by a negative number, the order relationship reverses. Think of it like this: $2 -5$, which is also true. If you didn't flip the sign, you would get $-2 < -5$, which is false. Always keep this in mind when isolating variables!

Q: Is it possible for the discriminant $D$ to be negative?

In this specific problem, we are dealing with the absolute difference of roots $|\beta - \alpha|$. If $D < 0$, the roots would be complex, and the difference would be imaginary, not a real number. Since the problem provides a real-valued range for the difference, we implicitly assume $D \ge 0$. Always check the nature of the roots when you see such constraints!

Question

Let $\alpha, \beta$ be the roots of the quadratic equation $12x^2 - 20x + 3\lambda = 0, \lambda \in \mathbb{Z}$. If $\frac{1}{2} \le |\beta - \alpha| \le \frac{3}{2}$, then the sum of all possible values of $\lambda$ is :

Options:

(A) 1

(B) 6

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a graph of a quadratic function, $f(x) = 12x^2 - 20x + 3\lambda$. This is a parabola, and the points where it touches the $x$-axis—the roots $\alpha$ and $\beta$—are the keys to unlocking the mystery of $\lambda$.

Our mission is to find the integer values of $\lambda$ that satisfy the geometric constraint: the distance between these two roots must be between $\frac{1}{2}$ and $\frac{3}{2}$.

### The Power of the Discriminant

To find the distance between the roots, we utilize the discriminant, $D = b^2 - 4ac$. The distance between the roots is given by the elegant formula:

$$|\beta - \alpha| = \frac{\sqrt{D}}{|a|}$$

Extracting our coefficients, we have $a = 12$, $b = -20$, and $c = 3\lambda$. Substituting these into the discriminant formula:

$$D = (-20)^2 - 4(12)(3\lambda) = 400 - 144\lambda$$

Factoring out $16$, we obtain $D = 16(25 - 9\lambda)$.

### The Simplification

Plugging this back into our distance formula, we get:

$$|\beta - \alpha| = \frac{\sqrt{16(25 - 9\lambda)}}{12}$$

Since the square root of $16$ is $4$, this simplifies beautifully:

$$|\beta - \alpha| = \frac{4\sqrt{25 - 9\lambda}}{12} = \frac{\sqrt{25 - 9\lambda}}{3}$$

We have now reduced the quadratic problem into a manageable expression involving $\lambda$.

### The Inequality Dance

We are given the constraint $\frac{1}{2} \le |\beta - \alpha| \le \frac{3}{2}$. Substituting our expression, we have:

$$\frac{1}{2} \le \frac{\sqrt{25 - 9\lambda}}{3} \le \frac{3}{2}$$

Multiplying the entire inequality by $3$ yields:

$$\frac{3}{2} \le \sqrt{25 - 9\lambda} \le \frac{9}{2}$$

Squaring all sides—a valid operation since all terms are positive—we get:

$$\frac{9}{4} \le 25 - 9\lambda \le \frac{81}{4}$$

Subtracting $25$ (or $\frac{100}{4}$) from all sides, we arrive at:

$$\frac{9 - 100}{4} \le -9\lambda \le \frac{81 - 100}{4} \Rightarrow -\frac{91}{4} \le -9\lambda \le -\frac{19}{4}$$

### The Final Reveal

Dividing by $-9$ requires us to flip the inequality signs. This gives us:

$$\frac{19}{36} \le \lambda \le \frac{91}{36}$$

Converting these to decimals, we find $0.527... \le \lambda \le 2.527...$. Since $\lambda$ must be an integer, the only possible values for $\lambda$ are **$1$** and **$2$**.

The sum of these values is $1 + 2 = \mathbf{3}$.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the quadratic equation $12 x^{2} - 20 x + 3 λ = 0, λ \in Z$ . If $\frac{1}{2} \leq ∣ β - α ∣ \leq \frac{3}{2}$ , then the sum of all possible values of $λ$ is :

Select Answer:

Visualized Solution

Identify the Quadratic Equation

The Difference of Roots Formula

Calculate the Discriminant $D$

Simplify the Discriminant

Simplify $∣ β - α ∣$

Apply the Given Inequality

Clear the Denominator

Square the Inequality

Isolate the Lambda Term

Solve for Lambda

Identify Integer Values and Final Sum

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of the Discriminant

The Simplification

The Inequality Dance

The Final Reveal

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