Let a, b ∈ R be such that the equation ax² - 2bx + 15 = 0 has a repeated root α. If α and β are the roots of the equation x² - 2bx + 21 = 0, then α ² + β ² is equal to :

Solution

Maths sir · 4 steps

A repeated root gives you its square straight from the product of roots. Plug it into the second equation.

1. Repeated root facts. For ax² - 2bx + 15 = 0 with both roots equal to α, the sum gives 2α = 2b/a, so α = b/a. The product gives α² = 15/a. Equal roots also need a zero discriminant: 4b² - 60a = 0, so b² = 15a.
2. Use the second equation. α is also a root of x² - 2bx + 21 = 0, so α² - 2bα + 21 = 0. Here 2bα = 2b²/a = (2 × 15a)/a = 30.

   15/a - 30 + 21 = 0 ⇒ α² = 15/a = 9

3. Find beta. The product of the roots of the second equation is αβ = 21, so β² = 441/α² = 441/9 = 49. As a check, a = 5/3 and b = ± 5 give α = ± 3 and β = ± 7, with α + β = 2b = ± 10.
4. So. α² + β² = 9 + 49 = 58, which is option (b).

The idea to keep. When a quadratic has a repeated root α, use both facts together: α is half the sum of the roots, and α² is the product. That usually removes the unknown coefficients quickly.

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