Related Formula
For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$:
Sum of roots: α + β = -(b)/(a)$\alpha + \beta = -\frac{b}{a}$
Product of roots: αβ = (c)/(a)$\alpha\beta = \frac{c}{a}$
Core Logic
From the given equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$:
α + β = (λ + 3)/(λ)$$\alpha + \beta = \frac{\lambda + 3}{\lambda}$$
αβ = (3)/(λ)$$\alpha\beta = \frac{3}{\lambda}$$
We are given the condition:
(1)/(α) - (1)/(β) = (1)/(3) (β - α)/(αβ) = (1)/(3)$$\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} \implies \frac{\beta - \alpha}{\alpha\beta} = \frac{1}{3}$$
β - α = (αβ)/(3) = (3/λ)/(3) = (1)/(λ)$$ \beta - \alpha = \frac{\alpha\beta}{3} = \frac{3/\lambda}{3} = \frac{1}{\lambda} $$
Step 1: Square Identity
We know that (α + β)² - 4αβ = (β - α)²$(\alpha + \beta)^2 - 4\alpha\beta = (\beta - \alpha)^2$.
Substitute the known values:
(β - α)² = (1)/(λ²)$$(\beta - \alpha)^2 = \frac{1}{\lambda^2}$$
(α + β)² = ((λ + 3)²)/(λ²)$$(\alpha + \beta)^2 = \frac{(\lambda + 3)^2}{\lambda^2}$$
So, we substitute these into the identity:
((λ + 3)²)/(λ²) - 4((3)/(λ)) = (1)/(λ²)$$\frac{(\lambda + 3)^2}{\lambda^2} - 4\left(\frac{3}{\lambda}\right) = \frac{1}{\lambda^2}$$
Step 2: Solving for Lambda
Multiply the entire equation by λ²$\lambda^2$ (assuming λ ≠ 0$\lambda \neq 0$ because it's a quadratic leading coefficient):
(λ + 3)² - 12λ = 1$$(\lambda + 3)^2 - 12\lambda = 1$$
λ² + 6λ + 9 - 12λ - 1 = 0$$\lambda^2 + 6\lambda + 9 - 12\lambda - 1 = 0$$
λ² - 6λ + 8 = 0$$\lambda^2 - 6\lambda + 8 = 0$$
Step 3: Finding Roots
(λ - 2)(λ - 4) = 0$$(\lambda - 2)(\lambda - 4) = 0$$
So, λ = 2$\lambda = 2$ or λ = 4$\lambda = 4$.
(The solution notes λ=0$\lambda=0$ derived from λ³ - 6λ² + 8λ = 0$\lambda^3 - 6\lambda^2 + 8\lambda = 0$ by cross multiplication, but λ=0$\lambda=0$ degrades the quadratic. Thus valid λ in 2, 4$\lambda \in \{2, 4\}$).
Sum of possible values of λ = 2 + 4 = 6$\lambda = 2 + 4 = 6$.
Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations