If α, β, where α

Q: Why can't we just solve for $\alpha$ and $\beta$ directly using the quadratic formula?

While you technically could, it would be a nightmare of radicals! The beauty of Vieta's formulas is that they allow us to manipulate the relationships between roots without ever needing to know their individual values. Always look for the symmetric properties first.

Q: Wait, why do we need to check the condition $\alpha < \beta$?

Great catch! The problem explicitly states $\alpha < \beta$. If our calculated $\lambda$ values resulted in complex roots or roots where $\alpha = \beta$ (meaning the discriminant is zero), those values would be invalid. We check to ensure our solutions are physically and mathematically consistent with the problem's constraints.

Q: Is $\lambda = 0$ a possible solution?

No, absolutely not! Look at the original equation: $\lambda x^2 - (\lambda + 3)x + 3 = 0$. If $\lambda$ were zero, the $x^2$ term would vanish, and the equation would no longer be quadratic. Since we are dealing with roots $\alpha$ and $\beta$, we must have a quadratic equation, so $\lambda$ must be non-zero.

Question

If $\alpha, \beta$, where $\alpha < \beta$, are the roots of the equation $\lambda x^2 - (\lambda + 3)x + 3 = 0$ such that $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of $\lambda$ is

Options:

(A) 8

(B) 6

(C) 2

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the precipice of a mathematical landscape, looking at the equation $\lambda x^2 - (\lambda + 3)x + 3 = 0$. At first glance, it looks like a standard quadratic, but it hides a secret.

We are given a condition on its roots, $\alpha$ and $\beta$, specifically that $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, with the constraint that $\alpha < \beta$. Our mission is to uncover the values of $\lambda$ that make this possible.

### The Power of Vieta's Formulas

Whenever we face a quadratic equation, our first instinct should be to invoke the wisdom of Vieta. We know that for any quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product is $c/a$.

Applying this to our equation, we find:

$$\alpha + \beta = \frac{\lambda + 3}{\lambda}$$

$$\alpha\beta = \frac{3}{\lambda}$$

These two expressions are the keys to the kingdom. They encapsulate everything we need to know about the roots without ever needing to solve for them individually.

### The Algebraic Bridge

Now, let's look at the condition $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$. By taking a common denominator, we get:

$$\frac{\beta - \alpha}{\alpha\beta} = \frac{1}{3}$$

This is a breakthrough! It tells us that the difference of the roots, $\beta - \alpha$, is directly proportional to the product of the roots.

Substituting our product $\alpha\beta = \frac{3}{\lambda}$, we find that $\beta - \alpha = \frac{1}{\lambda}$.

### The Dance of $\lambda$

We now have the sum, the product, and the difference of the roots. We use the fundamental identity: $(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.

Substituting our known values, we get:

$$\left( \frac{1}{\lambda} \right)^2 = \left( \frac{\lambda + 3}{\lambda} \right)^2 - 4 \left( \frac{3}{\lambda} \right)$$

Multiplying through by $\lambda^2$ (which we know is non-zero), we get $1 = (\lambda + 3)^2 - 12\lambda$. Expanding this, we arrive at $1 = \lambda^2 + 6\lambda + 9 - 12\lambda$, which simplifies beautifully to:

$$\lambda^2 - 6\lambda + 8 = 0$$

### Final Verification

Factorizing this quadratic is a joy: $(\lambda - 2)(\lambda - 4) = 0$. This gives us two candidates: $\lambda = 2$ and $\lambda = 4$.

For $\lambda = 2$, the equation becomes $2x^2 - 5x + 3 = 0$, with roots $1$ and $1.5$. Since $1 < 1.5$, this is valid.

For $\lambda = 4$, the equation becomes $4x^2 - 7x + 3 = 0$, with roots $0.75$ and $1$. Since $0.75 < 1$, this is also valid.

Both values hold up under scrutiny! The sum of all possible values of $\lambda$ is $2 + 4 = \mathbf{6}$.

Animated Solution for Mathematics - Quadratic Equations: If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is

Select Answer:

Visualized Solution

Given Equation and Condition

Sum and Product of Roots

Rearranging the Condition

Finding $β - α$

Connecting Difference, Sum, and Product

Substituting Expressions

Clearing the Fractions

Forming the Quadratic in $λ$

Factorizing the Equation

Checking the Validity of $λ$

Sum of All Possible Values

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Vieta's Formulas

The Algebraic Bridge

The Dance of $λ$

Final Verification

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