Multiple choice

If the roots of the quadratic equation $x^2 +(a + 6)x + b = 0$ are real and distinct for all $a \in R$ then the range of values of b is

  1. $(-\infty, \infty )$
  2. $(-\infty, 0 )$
  3. $(-6, \infty )$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be real and distinct, the discriminant D > 0. D = (a+6)^2 - 4b > 0. This must hold for all a in R. Since (a+6)^2 >= 0, the minimum value of (a+6)^2 is 0. For the inequality to hold for all a, we need -4b > 0, so b < 0.

AI explanation

For the roots to be real and distinct for all real values of a, the discriminant $D = (a+6)^2 - 4b$ must be strictly greater than zero. Since $(a+6)^2$ is always non-negative and reaches a minimum value of 0 when a is -6, substituting this minimum gives $0 - 4b > 0$. Solving this inequality yields b < 0. The range of values for b is (-infinity, 0).

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