Finding Lambda Values for Quadratic Roots

We are given the quadratic equation: $ \lambda x^2 - (\lambda + 3)x + 3 = 0 $ Let the roots of this equation be $\alpha$ and $\beta$, with the condition $\alpha < \beta$. We are also given the condition $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$.

Applying Vieta's Formulas

Using Vieta's formulas for the sum and product of the roots:

• Sum of roots: $\alpha + \beta = - \frac{-(\lambda + 3)}{\lambda} = \frac{\lambda + 3}{\lambda}$
• Product of roots: $\alpha \beta = \frac{3}{\lambda}$

For the formulas to be valid, we must have $\lambda \neq 0$. Also, for $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ to be defined, roots must be non-zero. Since $\alpha \beta = \frac{3}{\lambda}$, the roots are non-zero if $\lambda \neq 0$.

Manipulating the Root Condition

The given condition is $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$.

Combine the terms on the left side:

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

Substitute the expression for $\alpha \beta$ from Vieta's formulas:

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

Simplify the left side:

$ \frac{(\beta - \alpha)\lambda}{3} = \frac{1}{3} $

Multiply both sides by 3:

$ (\beta - \alpha)\lambda = 1 $

Isolate the difference of the roots:

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

Since $\alpha < \beta$, we know $\beta - \alpha > 0$. This implies $\frac{1}{\lambda} > 0$, which means $\lambda$ must be positive ($\lambda > 0$).

Relating Root Difference to Sum and Product

We use the identity $(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.

Substitute the expressions for $\alpha + \beta$, $\alpha \beta$, and $\beta - \alpha$:

$ \left(\frac{1}{\lambda}\right)^2 = \left(\frac{\lambda + 3}{\lambda}\right)^2 - 4\left(\frac{3}{\lambda}\right) $ $ \frac{1}{\lambda^2} = \frac{(\lambda + 3)^2}{\lambda^2} - \frac{12}{\lambda} $

To eliminate the denominators, multiply the entire equation by $\lambda^2$ (since $\lambda \neq 0$):

$ 1 = (\lambda + 3)^2 - 12\lambda $ $ 1 = (\lambda^2 + 6\lambda + 9) - 12\lambda $ $ 1 = \lambda^2 - 6\lambda + 9 $

Rearrange into a standard quadratic equation for $\lambda$:

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

Solving for Possible Lambda Values

Factor the quadratic equation:

$ (\lambda - 2)(\lambda - 4) = 0 $

This gives two possible values for $\lambda$:

$ \lambda = 2 \quad \text{or} \quad \lambda = 4 $

Both values are positive, satisfying the condition $\lambda > 0$. We also need the roots to be distinct, which requires the discriminant $\Delta = (\lambda-3)^2 > 0$, meaning $\lambda \neq 3$. Both $\lambda=2$ and $\lambda=4$ satisfy this.

Calculating the Sum of Possible Lambda Values

The possible values for $\lambda$ are $2$ and $4$.

The sum of all possible values of $\lambda$ is:

$ 2 + 4 = 6 $