Let α, β be the roots of the equation x² - px + r = 0 and α/2, 2β be the roots of the equation x² -

Q: Wait, why can't I just use the quadratic formula to find $x$ in terms of $p$ and $r$?

That is a very natural instinct! However, if you use the quadratic formula, you will introduce square roots (the discriminant) that make the algebra incredibly messy. Vieta's formulas are the 'shortcut' that bypasses the need to know the exact values of the roots. We only care about their sum and product, which is exactly what the coefficients provide. It is the elegant path, not the brute-force one.

Q: Why is the product of the roots the same for both equations?

Look closely at the constant term in both equations: $x^2 - px + r = 0$ and $x^2 - qx + r = 0$. The constant term is $r$ in both cases. Since the product of roots for a monic quadratic equation $x^2 - bx + c = 0$ is simply the constant term $c$, the product must be $r$ in both instances. This is a crucial observation that links the two equations together.

Q: Is there a specific order to solve for $\alpha$ and $\beta$?

Not at all! You can solve for $\beta$ first if you prefer. The goal is simply to express both $\alpha$ and $\beta$ in terms of $p$ and $q$. Once you have both, finding their product $r$ is just a matter of multiplication. Choose the path that feels most comfortable for your algebraic flow.

Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - px + r = 0$ and $\alpha/2, 2\beta$ be the roots of the equation $x^2 - qx + r = 0$. Then the value of $r$ is

Options:

(A) $\frac{2}{9} (p-q)(2q-p)$

(B) $\frac{2}{9} (q-p)(2p-q)$

(C) $\frac{2}{9} (q-2p)(2q-p)$

(D) $\frac{2}{9} (2p-q)(2q-p)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a complex algebraic puzzle involving two quadratic equations: $x^2 - px + r = 0$ and $x^2 - qx + r = 0$. At first glance, they appear as standard textbook problems, but they are connected by a shared constant $r$ and a specific transformation of roots.

Many students panic when they see roots like $\alpha/2$ and $2\beta$ and immediately reach for the quadratic formula. However, in JEE Advanced, the most powerful tool is often the one that avoids messy calculations: **Vieta's Formulas**.

### The First Foundation

Let us start with the first equation: $x^2 - px + r = 0$. We are told its roots are $\alpha$ and $\beta$.

Vieta's formulas act as the bridge between the roots and the coefficients. They state that the sum of the roots is the negative of the coefficient of $x$, and the product is the constant term.

We immediately write:
$$\alpha + \beta = p$$
$$\alpha\beta = r$$

These are our building blocks. Do not underestimate them, as they contain the entire DNA of the equation.

### The Transformation

Now, consider the second equation: $x^2 - qx + r = 0$. The roots are given as $\frac{\alpha}{2}$ and $2\beta$.

Applying Vieta's formulas again, the sum of these new roots is:
$$\frac{\alpha}{2} + 2\beta = q$$

The product of these roots is:
$$\left(\frac{\alpha}{2}\right)(2\beta) = \alpha\beta = r$$

Notice the beauty here: the product of the roots remains $\alpha\beta$, which is equal to $r$. This consistency confirms that our two equations are locked together by the same constant $r$.

### The Algebraic Dance

We now have a system of two linear equations:
1) $\alpha + \beta = p$
2) $\frac{\alpha}{2} + 2\beta = q$

Our goal is to find $r$, which is $\alpha\beta$. To get there, we must isolate $\alpha$ and $\beta$.

If we multiply the first equation by $2$, we get:
$$2\alpha + 2\beta = 2p$$

Now, subtract the second equation from this result:
$$\left(2\alpha - \frac{\alpha}{2}\right) + (2\beta - 2\beta) = 2p - q$$

The $\beta$ terms vanish, leaving us with:
$$\frac{3}{2}\alpha = 2p - q \implies \alpha = \frac{2}{3}(2p - q)$$

### The Final Convergence

With $\alpha$ in hand, finding $\beta$ is a simple matter of substitution. Using $\beta = p - \alpha$, we substitute our expression for $\alpha$:
$$\beta = p - \frac{2}{3}(2p - q) = \frac{3p - 4p + 2q}{3} = \frac{2q - p}{3}$$

Now, we have both roots expressed in terms of $p$ and $q$. The final step is to calculate $r = \alpha\beta$:
$$r = \left[ \frac{2}{3}(2p - q) \right] \cdot \left[ \frac{2q - p}{3} \right]$$

The denominators multiply to $9$, and the numerators yield the final result:
**$$r = \frac{2}{9}(2p - q)(2q - p)$$**

### Conclusion

We achieved this result without solving for $x$ or using the quadratic formula. We utilized the symmetry of the roots and the power of linear systems.

This is the essence of JEE Advanced mathematics: finding the path of least resistance through the forest of variables. Whenever you see roots of polynomials, think of Vieta, think of symmetry, and trust the process.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} - p x + r = 0$ and $α /2, 2 β$ be the roots of the equation $x^{2} - q x + r = 0$ . Then the value of $r$ is

Select Answer:

Visualized Solution

Analyze the First Equation $x^{2} - p x + r = 0$

Analyze the Second Equation $x^{2} - q x + r = 0$

Form the System of Equations

Eliminate $β$ to find $α$

Solve for $β$

Calculate the Product $r = α β$

Conclusion and Key Takeaways

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The First Foundation

The Transformation

The Algebraic Dance

The Final Convergence

Conclusion

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Visualized Solution

Analyze the First Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−px+r=0

Analyze the Second Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−qx+r=0