Let λ ≠ 0 be a real number. Let α, β be the roots of the equation 14x² - 31x + 3λ = 0 and α, γ be th

Q: Wait, why do we reject the idea of solving for $\lambda$ first? It seems like the most direct path.

That is a classic trap! While you *could* solve for $\lambda$, it introduces unnecessary complexity and potential for arithmetic errors. In JEE Advanced, the goal is often to find the most elegant path. By taking the ratio of the products, $\lambda$ cancels out immediately, simplifying the problem significantly. Always look for ways to eliminate parameters before diving into heavy calculation.

Q: How do I know when to add or subtract the equations?

Look at the structure of your equations. We have $\alpha + \beta$ and $\alpha + \gamma$. If you add them, you get $2\alpha + \beta + \gamma$, which doesn't help much. But if you subtract them, the common term $\alpha$ vanishes, leaving you with $\beta - \gamma$. Always aim to eliminate the common variable to isolate the unknowns.

Q: Is it really necessary to find the exact values of $\alpha, \beta,$ and $\gamma$?

In this specific problem, yes, it is the most robust method. Once you have the linear relationship between $\beta$ and $\gamma$ and the difference $\beta - \gamma$, the system becomes trivial to solve. Finding the roots ensures you have the exact values needed to construct the final quadratic equation without any ambiguity.

Question

Let $\lambda 
eq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 x^{2}-31 x+3 \lambda=0$ and $\alpha, \gamma$ be the roots of the equation $35 x^{2}-53 x+4 \lambda=0$. Then $\frac{3 \alpha}{\beta}$ and $\frac{4 \alpha}{\gamma}$ are the roots of the equation :

Options:

(A) 7 x^{2}+245 x-250=0

(B) 7 x^{2}-245 x+250=0

(C) 49 x^{2}-245 x+250=0

(D) 49 x^{2}+245 x+250=0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before two seemingly independent quadratic equations: 
$$14x^2 - 31x + 3\lambda = 0$$ 
and 
$$35x^2 - 53x + 4\lambda = 0$$

At first glance, they look like two separate problems, but there is a hidden connection—a common root, $\alpha$. This $\alpha$ is the key that unlocks the entire structure. In the world of JEE Advanced, recognizing this commonality is your first step toward victory.

### The Vieta's Toolkit

We begin by invoking the legendary Vieta's formulas. For the first equation, the sum of the roots is:
$$\alpha + \beta = \frac{31}{14}$$ 
and the product is:
$$\alpha\beta = \frac{3\lambda}{14}$$

For the second equation, we have:
$$\alpha + \gamma = \frac{53}{35}$$ 
and 
$$\alpha\gamma = \frac{4\lambda}{35}$$

These are our building blocks. Do not be intimidated by the parameter $\lambda$; it is a ghost in the machine, and we are about to exorcise it.

### The Art of Elimination

Why carry $\lambda$ through the entire calculation? It is unnecessary weight. By taking the ratio of the products:
$$\frac{\alpha\beta}{\alpha\gamma} = \frac{3\lambda/14}{4\lambda/35}$$

The $\lambda$ and $\alpha$ terms cancel out, leaving us with:
$$\frac{\beta}{\gamma} = \frac{3}{14} \cdot \frac{35}{4} = \frac{15}{8}$$

This gives us a beautiful, simple relationship: $\beta = \frac{15}{8}\gamma$.

Now, look at the sums. If we subtract the second sum from the first:
$$(\alpha + \beta) - (\alpha + \gamma) = \frac{31}{14} - \frac{53}{35}$$

The common root $\alpha$ disappears entirely! The result is:
$$\beta - \gamma = \frac{7}{10}$$

We now have a system of two linear equations with two variables. Substituting $\beta = \frac{15}{8}\gamma$ into $\beta - \gamma = \frac{7}{10}$ gives us:
$$\frac{7}{8}\gamma = \frac{7}{10}$$

This simplifies to $\gamma = \frac{4}{5}$. From here, $\beta$ falls into place as $\frac{3}{2}$, and $\alpha$ is revealed to be $\frac{5}{7}$.

### The Final Construction

With $\alpha$, $\beta$, and $\gamma$ in hand, the rest is a victory lap. The problem asks for a new quadratic equation with roots $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$.

Calculating these, we find the roots are:
$$\frac{3(5/7)}{3/2} = \frac{10}{7} \quad 	ext{and} \quad \frac{4(5/7)}{4/5} = \frac{25}{7}$$

The sum of these roots is $S = 5$, and the product is $P = \frac{250}{49}$. Using the standard form $x^2 - Sx + P = 0$, we arrive at:
$$x^2 - 5x + \frac{250}{49} = 0$$

Multiplying by $49$ gives us the final, elegant result:
**$49x^2 - 245x + 250 = 0$**

You have successfully navigated the complexity and emerged with the correct answer. This is the power of systematic, logical thinking in mathematics.

Select Answer:

Visualized Solution

Given Equations and Roots

Sum and Product of Roots for Eq 1

Sum and Product of Roots for Eq 2

Ratio of Products

Difference of Sums

Calculating $γ$

Calculating $β$

Calculating $α$

Calculating the Required Roots

Sum and Product for New Equation

Final Quadratic Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Vieta's Toolkit

The Art of Elimination

The Final Construction

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