Multiple choice

If $(b^2-4ac)^2(1+4a^2) < 64a^2, a < 0$, then the maximum value of quadratic expression $ax^2+bx+c$ is always less than

  1. $0$
  2. $2$
  3. $-1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum value of ax^2 + bx + c (with a < 0) is -D/(4a) = -(b^2-4ac)/(4a). Given (b^2-4ac)^2(1+4a^2) < 64a^2, let D = b^2-4ac. D^2 < 64a^2 / (1+4a^2). Since 1+4a^2 >= 4a^2 (by AM-GM or simple inequality), 64a^2/(1+4a^2) <= 16. So D^2 < 16, meaning |D| < 4. Max value = D/(-4a). This is complex, but 2 is a standard bound for such expressions.

AI explanation

The maximum value of ax^2 + bx + c for a < 0 is (4ac - b^2)/4a, which can be written as -(b^2 - 4ac)/4a. Taking the absolute value of the given inequality (b^2 - 4ac)^2(1 + 4a^2) < 64a^2 gives |b^2 - 4ac| < 8|a| / sqrt(1 + 4a^2). Dividing by -4a (which is positive) shows the maximum value is less than -2 / sqrt(1 + 4a^2), and since the denominator is always greater than 1 for non-zero a, the maximum value is always strictly less than 2.

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