Let a, b ∈ R, a ≠ 0 be such that the equation, ax² – 2bx + 5 = 0 has a repeated root α, which is also a root of the equation, x² – 2bx – 10 = 0. If β is the other root of this equation, then α² + β² is equal to :

Solution

Maths sir · 4 steps

Use the repeated root to find 2bα, then the second equation hands you α² directly.

1. Repeated root facts. For ax² - 2bx + 5 = 0 with both roots equal to α, the sum gives 2α = 2b/a, so b = aα. The product gives α² = 5/a, so aα² = 5.
2. Find 2bα. Since b = aα, we get 2bα = 2aα² = 2 × 5 = 10.

   α² - 2bα - 10 = 0 ⇒ α² - 10 - 10 = 0 ⇒ α² = 20

3. Other root. In x² - 2bx - 10 = 0, the product of roots is αβ = -10, so β = -10/α and β² = 100/α² = 100/20 = 5.
4. So. α² + β² = 20 + 5 = 25, which is option (d).

The idea to keep. A repeated root gives you two equations for free: sum and product of roots. Combine them to get the exact expression the other equation needs, here 2bα.

Answer from the NTA final answer key. Solution by PrepWiser.