Multiple choice

Find all values of parameter $a$ for which the quadratic equation $ \left( a+1 \right) { x }^{ 2 }+2\left( a+1 \right) x+a-2=0 $ has two distinct roots.

  1. $a \in (-1,\infty)$
  2. $a \in (-\infty,-1)$
  3. $a \in (-1,1)$
  4. $a \in (-\infty,\infty)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two distinct roots, the discriminant D > 0 and the coefficient of x^2 must not be zero. D = (2(a+1))^2 - 4(a+1)(a-2) = 4(a+1)^2 - 4(a+1)(a-2) = 4(a+1)(a+1 - (a-2)) = 4(a+1)(3) = 12(a+1). 12(a+1) > 0 => a > -1. Also a+1 != 0 => a != -1.

AI explanation

For the equation (a + 1)x^2 + 2(a + 1)x + a - 2 = 0 to have two distinct real roots, the discriminant must be positive, so [2(a + 1)]^2 - 4(a + 1)(a - 2) > 0. This simplifies to 4(a + 1)[(a + 1) - (a - 2)] > 0, which further reduces to 4(a + 1)(3) > 0. Solving the inequality 12(a + 1) > 0 yields a > -1, so a is in the interval (-1, infinity).

Keep practicing — related questions