Let α and β be the roots of equation px² + qx + r = 0, p ≠ 0. If p, q, r are in A.P. and 1/α + 1/β =

Q: Wait, why can't we just find the roots $\alpha$ and $\beta$ individually using the quadratic formula?

That is a great instinct, but it is a trap! If you try to solve for $\alpha$ and $\beta$ directly, you will end up with a massive, messy expression involving square roots of $p, q,$ and $r$. In JEE Advanced, the goal is often to find relationships between roots without ever knowing the roots themselves. By using symmetric functions like the sum and product, we bypass the messy algebra and get straight to the elegant solution.

Q: Why does the $|r|$ cancel out in the final step? Is it possible for $r$ to be zero?

Excellent observation! If $r$ were zero, the equation would become $px^2 + qx = 0$, which means one root is zero. But the problem gives us $\frac{1}{\alpha} + \frac{1}{\beta} = 4$, which implies $\alpha$ and $\beta$ cannot be zero. Therefore, $r$ cannot be zero. Since $r \neq 0$, we are mathematically safe to cancel $|r|$ from the numerator and denominator.

Q: Is there a faster way to find $|\alpha - \beta|$ without the discriminant formula?

Yes, there is! You can use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$. This is essentially the same as the discriminant method, just written differently. It is a powerful tool to keep in your arsenal because it relies purely on the sum and product of roots, which we already found.

Question

Let $\alpha$ and $\beta$ be the roots of equation $px^2 + qx + r = 0, p 
eq 0$. If $p, q, r$ are in A.P. and $1/\alpha + 1/\beta = 4$, then the value of $|\alpha - \beta|$ is:

Options:

(A) $\sqrt{34} / 9$

(B) 2\sqrt{13} / 9$

(C) $\sqrt{61} / 9$

(D) 2\sqrt{17} / 9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a quadratic equation; we are uncovering the hidden architecture of numbers.

When you look at a quadratic equation like $px^2 + qx + r = 0$, do not just see a string of symbols. See a story where the roots $\alpha$ and $\beta$ are two points dancing on the number line, their positions dictated by the coefficients $p, q,$ and $r$.

Our mission is to find the distance between them, $|\alpha - \beta|$, using only the clues provided.

### The Master Keys

Every quadratic equation carries its own DNA, encoded in the relationship between its roots and its coefficients. We call these **Vieta's Formulas**.

For our equation $px^2 + qx + r = 0$, the sum and product of the roots are:

$$\alpha + \beta = -\frac{q}{p}$$

$$\alpha\beta = \frac{r}{p}$$

These two simple relations are the foundation upon which we will build our entire solution. Never underestimate the power of these identities; they are the most reliable tools in your JEE toolkit.

### The Algebraic Alchemy

We are given a curious condition: $\frac{1}{\alpha} + \frac{1}{\beta} = 4$. By taking the lowest common multiple, we obtain:

$$\frac{\alpha + \beta}{\alpha\beta} = 4$$

Suddenly, the fog clears! We have the sum of the roots in the numerator and the product in the denominator. By substituting our master keys, we get:

$$\frac{-\frac{q}{p}}{\frac{r}{p}} = 4$$

Notice the beauty of the cancellation—the $p$ vanishes, leaving us with $-\frac{q}{r} = 4$, or simply $q = -4r$. We have just reduced the complexity of our problem by linking two of our coefficients together.

### The A.P. Bridge

Now, the problem introduces a new constraint: $p, q,$ and $r$ are in an Arithmetic Progression (A.P.). In an A.P., the middle term is the average of the neighbors, which means:

$$2q = p + r$$

We now have two equations: $q = -4r$ and $2q = p + r$. By substituting the first into the second, we get:

$$2(-4r) = p + r \Rightarrow -8r = p + r \Rightarrow p = -9r$$

We have now expressed both $p$ and $q$ in terms of $r$. We have successfully tamed the variables!

### The Final Descent

We are ready for the final act. We need to find $|\alpha - \beta|$. The direct formula for the difference of roots is:

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

Where $D = q^2 - 4pr$ is the discriminant. Let's plug in our values:

$$D = (-4r)^2 - 4(-9r)(r) = 16r^2 + 36r^2 = 52r^2$$

Now, the difference is:

$$|\alpha - \beta| = \frac{\sqrt{52r^2}}{|p|} = \frac{\sqrt{52}|r|}{|-9r|}$$

The $|r|$ terms cancel out beautifully, leaving us with $\frac{\sqrt{52}}{9}$. Since $\sqrt{52} = \sqrt{4 	imes 13} = 2\sqrt{13}$, our final answer is:

$$\mathbf{\frac{2\sqrt{13}}{9}}$$

Take a moment to appreciate this. We started with a general quadratic equation and a few constraints, and through logical deduction, we arrived at a precise, elegant constant. This is the thrill of JEE mathematics.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the roots of equation $p x^{2} + q x + r = 0, p \neq = 0$ . If $p, q, r$ are in A.P. and $1/ α + 1/ β = 4$ , then the value of $∣ α - β ∣$ is:

Select Answer:

Visualized Solution

The Given Equation

Sum and Product of Roots

Simplifying the Given Condition

Relate $q$ and $r$

The A.P. Condition

Solve for $p$ in terms of $r$

Formula for $∣ α - β ∣$

Substitute $p$ and $q$

Simplify the Discriminant

Final Calculation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Master Keys

The Algebraic Alchemy

The A.P

The Final Descent

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