Multiple choice

If $\displaystyle x^{2}+(a-b)x+(1-a-b)= 0, $ where $\displaystyle a,b \in R,$ the value of $a$ such that the equation has distinct real roots for all value of $b$ are

  1. $\displaystyle a> 1 $
  2. $\displaystyle a< 1 $
  3. $\displaystyle a> 2 $
  4. $\displaystyle a< 2 $
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A Correct answer
Explanation

For distinct real roots, the discriminant D = (a-b)^2 - 4(1-a-b) > 0. Expanding: a^2 - 2ab + b^2 - 4 + 4a + 4b > 0. This must hold for all b, which implies the quadratic in b must have a negative discriminant.

AI explanation

For the quadratic equation x^2 + (a-b)x + (1-a-b) = 0 to have distinct real roots, its discriminant must be strictly positive for all real values of b. The discriminant is D = (a-b)^2 - 4(1)(1-a-b) = a^2 - 2ab + b^2 - 4 + 4a + 4b = b^2 + (4 - 2a)b + (a^2 + 4a - 4). For this quadratic expression in b to be positive for all b, its own discriminant must be negative, meaning (4 - 2a)^2 - 4(1)(a^2 + 4a - 4) < 0. Simplifying gives 16 - 16a + 4a^2 - 4a^2 - 16a + 16 < 0, which reduces to 32 - 32a < 0. Dividing by 32 yields a > 1.

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