Multiple choice

If the equation $ax^2+2bx-3c=0$ has non real roots and $(3 c/4) < (a+b)$; then $C$ is always

  1. $< 0$
  2. $> 0$
  3. $\ge 0$

  4. Zero
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A Correct answer
Explanation

For non-real roots, the discriminant D = (2b)^2 - 4(a)(-3c) = 4b^2 + 12ac < 0, so b^2 + 3ac < 0. Given 3c/4 < a + b, this implies 3c < 4a + 4b. Analysis of the quadratic properties confirms c must be negative for these conditions to hold simultaneously.

AI explanation

Because the quadratic equation ax^2 + 2bx - 3c = 0 has non-real roots, its discriminant must be negative, giving (2b)^2 - 4(a)(-3c) < 0 or b^2 + 3ac < 0. Since b^2 is non-negative, 3ac must be strictly negative, meaning ac < 0. Rearranging the given inequality 3c/4 < a + b yields 3c < 4a + 4b, which combined with the condition ac < 0 allows us to conclude that c is always negative.

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