Multiple choice

If the equation $a{x^2} + bx + 6 = 0$ does not have two distinct real roots, then the least value of $3a + b$ is

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

For the quadratic ax^2 + bx + 6 = 0 to not have two distinct real roots, the discriminant D = b^2 - 4(a)(6) must be <= 0. So b^2 <= 24a. We want the minimum value of 3a + b. Testing values, if a=1, b^2 <= 24, so b can be -4. 3(1) - 4 = -1. If a=1/3, b^2 <= 8, b approx -2.8. If a=1/6, b^2 <= 4, b = -2. 3(1/6) - 2 = 0.5 - 2 = -1.5. Checking the option -2: if 3a+b = -2, b = -2-3a. (-2-3a)^2 <= 24a -> 4 + 12a + 9a^2 <= 24a -> 9a^2 - 12a + 4 <= 0 -> (3a-2)^2 <= 0. This holds for a=2/3, b=-4.

AI explanation

For the equation to not have two distinct real roots, the discriminant b^2 - 24a must be less than or equal to zero, meaning b^2 is less than or equal to 24a. This requires a to be positive. To minimize 3a + b, we take the extreme case b = -sqrt(24a). Substituting this gives 3a - sqrt(24a), which is minimized when a = 2/3. At this value, b = -4, resulting in a minimum value of -2 for the expression.

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