Aditi Chauhan
Last Activity: 10 Years ago
Sol. The given equation is,
x2 + (a – b) x + (1 – a – b) = 0
a, b ∈ R
For the eqn to have unequal real roots ∀ b D > 0
⇒ (a – b)2 – 4 (1 – a – b) > 0
⇒ a2 + b2 – 2ab -4 + 4a + 4b > 0
⇒ b2 + b(4 – 2a) + a2 + 4a – 4 > 0
Which is a quadratic expression in b, and it will be true ∀ b ∈ R if discriminant of above of above eqn less than zero.
i.e., (4 – 2a)2 – 4(a2 + 4a – 4) < 0
⇒ (2 – a)2 – (a2 + 4a – 4) < 0
⇒ 4 – 4a + a2 – a2 – 4a + a < 0
⇒ - 8a + 8 < 0
⇒ a > 1
