Quadratic Equations Question - None

Question

Let the original equation have two roots $\alpha$ and $\beta$. Then $\alpha\beta = \alpha^2\beta^2$ ... (i) and $\alpha^2 + \beta^2 = \alpha + \beta$ ... (ii). Find the number of quadratic equations satisfying these conditions.

Accepted Answer

Step 1: From condition (i): αβ = α²β² ⟹ αβ(αβ - 1) = 0 ⟹ αβ = 0 or αβ = 1

Step 2: From condition (ii): α² + β² = α + β. Using α² + β² = (α + β)² - 2αβ, we get (α + β)² - 2αβ = α + β

Case 1 (αβ = 0): Let α = 0. Then β² = β ⟹ β = 0 or β = 1
• If β = 0: equation is x² = 0 (one equation)
• If β = 1: equation is x² - x = 0 (one equation)

Case 2 (αβ = 1): Substitute into condition (ii): (α + β)² - 2(1) = α + β
Let s = α + β: s² - s - 2 = 0 ⟹ s = 2 or s = -1
• If s = 2: α + β = 2, αβ = 1 ⟹ α, β are roots of t² - 2t + 1 = 0 ⟹ α = β = 1, giving equation x² - 2x + 1 = 0 (one equation)
• If s = -1: α + β = -1, αβ = 1 ⟹ α, β are roots of t² + t + 1 = 0 ⟹ roots are complex (valid if allowed, one equation)

Step 3: The distinct quadratic equations are:
1) x² = 0
2) x² - x = 0
3) x² - 2x + 1 = 0
4) x² + x + 1 = 0 (if complex roots allowed)

Note: If only real roots required: 3 equations. If complex allowed: 4 equations.

∴ Answer: C (typically 4, depending on context)

Question:

Step-by-Step Solution

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