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

Step-by-step solution

Given, a x 2 - 2 b x + 5 = 0 has repeated root α . ∴ 2 α = 2 b a ⇒ α = b a and α 2 = 5 a ⇒ b 2 a 2 = 5 a ⇒ b 2 = 5 a   . . . i a ≠ 0 α + β = 2 b   . . . ii and α β = - 10     . . . iii α = b a is also root of x 2 - 2 b x - 10 = 0 ⇒ b 2 - 2 a b 2 - 10 a 2 = 0 by i ⇒ 5 a - 10 a 2 - 10 a 2 = 0 ⇒ 20 a 2 = 5 a ⇒ a = 1 4 and b 2 = 5 4 Now α 2 + β 2 = α + β 2 - 2 α β = 5 + 20