Let λ ≠ 0 be in R. If α and β are the roots of the equation, x² - x + 2λ = 0 and α and γ are the roo

Q: Wait, why can't we just solve for $x$ using the quadratic formula? Wouldn't that be faster?

While the quadratic formula is a powerful tool, it would introduce square roots involving $\lambda$, making the algebra significantly more complex and prone to errors. Using Vieta's formulas allows us to work with the relationships between roots and coefficients directly, which is the elegant, intended path for this type of problem.

Q: Why did we subtract the two product equations in the final steps?

We subtracted them specifically to eliminate the $\alpha^2$ term. By doing so, we reduced a system of quadratic equations into a simple linear relationship between $\alpha$ and $\lambda$, which is much easier to solve.

Q: What if $\lambda = 0$? The problem says $\lambda \ne 0$, but what if it were?

If $\lambda$ were zero, the equations would simplify to $x^2 - x = 0$ and $3x^2 - 10x = 0$. The roots would be $0$ and $1$ for the first, and $0$ and $10/3$ for the second. While $0$ is a common root, the expression $\frac{\beta\gamma}{\lambda}$ would become undefined (division by zero). The condition $\lambda \ne 0$ ensures the expression is well-defined.

Question

Let $\lambda 
e 0$ be in $R$. If $\alpha$ and $\beta$ are the roots of the equation, $x^2 - x + 2\lambda = 0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3x^2 - 10x + 27\lambda = 0$, then $\frac{\beta\gamma}{\lambda}$ is equal to :

Options:

(A) 18

(B) 9

(C) 27

(D) 36

Sigma Percentile · Accepted Answer

### The Mystery of the Common Root

In the realm of JEE Advanced mathematics, quadratic equations are not just algebraic expressions; they are geometric entities, stories of intersection, and puzzles of symmetry.

We are given two quadratic equations: 
$$x^2 - x + 2\lambda = 0$$ 
and 
$$3x^2 - 10x + 27\lambda = 0$$

They share a common root, $\alpha$. This common root is the key that unlocks the entire problem. When you see a common root, think of it as a bridge connecting two different worlds.

### Phase 1: The Vieta's Toolkit

Before we dive into the algebra, let us sharpen our tools. Vieta's formulas are our best friends here.

For the first equation, $x^2 - x + 2\lambda = 0$, with roots $\alpha$ and $\beta$, we know:
$$\alpha + \beta = 1$$
$$\alpha\beta = 2\lambda$$

For the second equation, $3x^2 - 10x + 27\lambda = 0$, with roots $\alpha$ and $\gamma$, we apply Vieta's again:
$$\alpha + \gamma = \frac{10}{3}$$
$$\alpha\gamma = \frac{27\lambda}{3} = 9\lambda$$

Notice how the common root $\alpha$ appears in both sets of equations. This is exactly what we need to exploit.

### Phase 2: The Algebraic Dance

We now have a system of equations. Our goal is to find the value of $\frac{\beta\gamma}{\lambda}$.

To get there, we express $\beta$ and $\gamma$ in terms of $\alpha$:
$$\beta = 1 - \alpha$$
$$\gamma = \frac{10}{3} - \alpha$$

Now, substitute these into our product relations:
$$\alpha(1 - \alpha) = 2\lambda \Rightarrow \alpha - \alpha^2 = 2\lambda$$
$$\alpha\left(\frac{10}{3} - \alphaight) = 9\lambda \Rightarrow \frac{10\alpha}{3} - \alpha^2 = 9\lambda$$

We have two equations, both containing an $-\alpha^2$ term. If we subtract the first from the second, the $\alpha^2$ terms will vanish into thin air.

### Phase 3: The Final Reveal

Subtracting the equations gives us:
$$\left(\frac{10\alpha}{3} - \alpha^2ight) - (\alpha - \alpha^2) = 9\lambda - 2\lambda$$

Simplifying this, we get:
$$\frac{7\alpha}{3} = 7\lambda \Rightarrow \alpha = 3\lambda$$

Now, substitute $\alpha = 3\lambda$ back into our first expanded equation:
$$3\lambda - (3\lambda)^2 = 2\lambda$$
$$3\lambda - 9\lambda^2 = 2\lambda \Rightarrow \lambda - 9\lambda^2 = 0$$

Since the problem guarantees $\lambda 
eq 0$, we divide by $\lambda$ to get $1 - 9\lambda = 0$, which means $\lambda = \frac{1}{9}$.

With $\lambda$ in hand, we find:
$$\alpha = 3\left(\frac{1}{9}ight) = \frac{1}{3}$$
$$\beta = 1 - \frac{1}{3} = \frac{2}{3}$$
$$\gamma = \frac{10}{3} - \frac{1}{3} = 3$$

Finally, the product $\beta\gamma = \left(\frac{2}{3}ight) 	imes 3 = 2$. Dividing by $\lambda$:
$$\frac{\beta\gamma}{\lambda} = \frac{2}{1/9} = 18$$

We have arrived at the answer, **18**, not by brute force, but by understanding the elegant structure of the problem.

Animated Solution for Mathematics - Quadratic Equations: Let $λ \neq = 0$ be in $R$ . If $α$ and $β$ are the roots of the equation, $x^{2} - x + 2 λ = 0$ and $α$ and $γ$ are the roots of the equation, $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to :

Select Answer:

Visualized Solution

First Quadratic Equation

Sum and Product of Roots for Equation 1

Second Quadratic Equation

Sum and Product of Roots for Equation 2

Isolating $β$ and $γ$

Setting up Equations in $α$ and $λ$

Solving for $α$ in terms of $λ$

Finding the Value of $λ$

Finding Specific Values of Roots

Evaluating $\frac{β γ}{λ}$

The Sigma Insight: Common Roots

The Mystery of the Common Root

Phase 1

Phase 2

Phase 3

Similar Questions

Let $λ \neq = 0$ be in $R$ . If $α$ and $β$ are the roots of the equation $x^{2} - x + 2 λ = 0$ , and $α$ and $γ$ are the roots of equation $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to .

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real roots is $\frac{3}{2 β}$ , then $β$ is equal to

If $α, β$ are the roots of $x^{2} + p x + q = 0$ and $γ, δ$ are the roots of $x^{2} + r x + s = 0$ , evaluate $(α - γ) (α - δ) (β - γ) (β - δ)$ in terms of $p, q, r$ and $s$ . Deduce the condition that the equations have a common root.

If the equations $x^{2} + 2 x + 3 = 0$ and $a x^{2} + b x + c = 0, a, b, c \in R$ , have a common root, then $a : b : c$ is

The quadritic equations $x^{2} - 6 x + a = 0$ and $x^{2} - c x + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

If $α, β$ and $γ$ are three consecutive terms of a non-constant G.P. such that the equations $x^{2} + 2 α x + β = 0$ and $x^{2} + β x - 1 = 0$ have a common root, then $α (β + γ)$ is equal to :

If the quadratic equations $x^{2} + a x + b = 0$ and $x^{2} + b x + a = 0$ ( $a \neq = b$ ) have a common root, then the numerical value of $a + b$ is .........

A value of $b$ for which the equations $x^{2} + b x - 1 = 0, x^{2} + x + b = 0$ have one root in common is

If $a, b, c$ are in G.P., then the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root if $d / a, e / b, f / c$ are in

STATEMENT-1: $(p^{2} - q) (b^{2} - a c) \geq 0$ and STATEMENT-2: $b \neq = p a$ or $c \neq = q a$

Select Answer:

Visualized Solution

First Quadratic Equation

Sum and Product of Roots for Equation 1

Second Quadratic Equation

Sum and Product of Roots for Equation 2

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ

Setting up Equations in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding Specific Values of Roots

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λβγ​