Multiple choice

If equation $P(x)=x^2+ax+1$ has two distinct real roots, then exhaustive values of $a$ are

  1. $(-2, 2)$
  2. $(-\infty, -2)\cup (2, \infty)$
  3. $(-2, \infty)$

  4. all real numbers
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B Correct answer
Explanation

For distinct real roots, the discriminant D > 0. D = a^2 - 4(1)(1) = a^2 - 4. a^2 - 4 > 0 implies a^2 > 4, so a > 2 or a < -2.

AI explanation

A quadratic equation has two distinct real roots when its discriminant is strictly greater than zero. For the equation x^2 + ax + 1 = 0, we set the discriminant D = a^2 - 4(1)(1) > 0. This gives the inequality a^2 > 4, which is satisfied when a is less than -2 or greater than 2. Therefore, the exhaustive values of a fall in the set (-infinity, -2) union (2, infinity).

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