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: Why do we subtract the equations? Can't we just use the quadratic formula?

You could use the quadratic formula, but that would lead to messy radicals and a much longer path. The 'common root' technique is a classic JEE strategy: by eliminating the highest-degree term (the $\alpha^2$ term), you reduce a complex quadratic system into a simple linear relationship. It is the most elegant and error-proof way to proceed.

Q: Why did we reject $\lambda = 0$?

Great catch! The problem explicitly states $\lambda \neq 0$. In competitive exams, always keep an eye on the constraints given in the question stem. If you had included $\lambda = 0$, you would have introduced an extraneous solution that violates the problem's definition.

Q: Is there a shortcut using the product of roots?

While you can use the product of roots ($\alpha\beta = 2\lambda$ and $\alpha\gamma = 9\lambda$), you still need the value of $\alpha$ to find $\beta$ and $\gamma$. Finding the common root first is the most reliable way to ensure you don't miss any steps or make algebraic errors.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the intersection of two mathematical paths. You have two quadratic equations:
$x^2 - x + 2\lambda = 0$ and $3x^2 - 10x + 27\lambda = 0$.

They share a secret: a common root, $\alpha$. This root is the key that unlocks the entire problem.

If $\alpha$ is a root of the first equation, it satisfies:
$$\alpha^2 - \alpha + 2\lambda = 0$$

If it is a root of the second, it satisfies:
$$3\alpha^2 - 10\alpha + 27\lambda = 0$$

### The Algebraic Dance: Eliminating the Obstacle

We have a system of two equations. Our goal is to find $\alpha$ and $\lambda$. The $\alpha^2$ term is the obstacle preventing us from finding a direct relationship.

We use the power of elimination. If we multiply the first equation by $3$, we get:
$$3\alpha^2 - 3\alpha + 6\lambda = 0$$

Now, consider the second equation:
$$3\alpha^2 - 10\alpha + 27\lambda = 0$$

By subtracting the first from the second, the $\alpha^2$ terms vanish. We are left with:
$$(-10\alpha - (-3\alpha)) + (27\lambda - 6\lambda) = 0$$

This simplifies beautifully to $-7\alpha + 21\lambda = 0$, which yields the linear relationship:
$$\alpha = 3\lambda$$

### Unlocking the Mystery of $\lambda$

Now that we know $\alpha = 3\lambda$, we substitute this into the first equation:
$$(3\lambda)^2 - (3\lambda) + 2\lambda = 0$$

This expands to:
$$9\lambda^2 - \lambda = 0$$

Factoring this, we get $\lambda(9\lambda - 1) = 0$. Since the problem explicitly states $\lambda 
eq 0$, we discard that root and find:
$$\lambda = \frac{1}{9}$$

### The Final Reveal

With $\lambda = \frac{1}{9}$, we find our common root:
$$\alpha = 3\lambda = 3\left(\frac{1}{9}ight) = \frac{1}{3}$$

For the first equation, the sum of roots is $\alpha + \beta = 1$. Since $\alpha = \frac{1}{3}$, then:
$$\beta = 1 - \frac{1}{3} = \frac{2}{3}$$

For the second equation, the sum of roots is $\alpha + \gamma = \frac{10}{3}$. Since $\alpha = \frac{1}{3}$, then:
$$\gamma = \frac{10}{3} - \frac{1}{3} = 3$$

Finally, we calculate the expression $\frac{\beta\gamma}{\lambda}$:
$$\frac{\left(\frac{2}{3} 	imes 3ight)}{\frac{1}{9}} = \frac{2}{\frac{1}{9}} = 18$$

The journey is complete. We have navigated the algebra and arrived at the elegant result of **18**.

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 equation $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to .

Enter Numerical Value:

Visualized Solution

Identifying the Common Root $α$

Setting up Equations for $α$

Eliminating the Quadratic Term $α^{2}$

Finding the Relation Between $α$ and $λ$

Substituting $α$ to Find $λ$

Solving for $λ$

Determining the Value of $α$

Finding the Root $β$

Finding the Root $γ$

Setting Up the Final Expression

Final Calculation of $\frac{β γ}{λ}$

Conclusion & Key Takeaway

The Sigma Insight: Common Roots

Analyzing the Setup

The Algebraic Dance

Unlocking the Mystery of $λ$

The Final Reveal

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 the equation, $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to :

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

Enter Numerical Value:

Visualized Solution

Identifying the Common Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

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

Eliminating the Quadratic Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2

Finding the Relation Between .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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α to Find .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; } λ

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

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

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