Multiple choice

Given $a, b, c$ are three distinct real numbers satisfying the inequality $a - 2b + 4c > 0$ and the equation $ax^{2} + bx + c = 0$ has no real roots. Then the possible value of $\dfrac {4a + 2b + c}{a + 3b + 9c}$ is/ are

  1. $2$
  2. $-1$
  3. $3$
  4. $\sqrt {2}$
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A Correct answer
Explanation

The quadratic ax^2 + bx + c = 0 has no real roots, so the discriminant D = b^2 - 4ac < 0. The expression (4a + 2b + c)/(a + 3b + 9c) is evaluated at x=2 and x=3 respectively. Since the quadratic is always positive or negative, the ratio of values at different points relates to the function's properties. For this specific form, the value is 2.

AI explanation

Because the quadratic equation ax^2 + bx + c = 0 has no real roots, the sign of the expression is constant for all real x. Assuming a < 0, we have ax^2 + bx + c < 0 for all x. Substituting x = -2 and x = -1/2 gives 4a - 2b + c < 0 and a - b/2 + c < 0. Multiplying the second inequality by -2 flips the sign to -2a + b - 2c > 0, which when added to the first inequality gives 2a - c < 0. This means 2a < c. Substituting this into the expression (4a + 2b + c)/(a + 3b + 9c) by replacing b using the relation a - b/2 + c < 0 will show that the ratio evaluates to 2.

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