Multiple choice

If $2{ a }+3{ b }+6{ c }=0$, then at least one root of the equation ${ a }{ x }^{ 2 }+bx+{ c }=0$ lies in the interval:

  1. $(0, 1)$
  2. $(1, 2)$
  3. $(2, 3)$
  4. $(1, 3)$
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A Correct answer
Explanation

Let f(x) = ax^2 + bx + c. We are given 2a + 3b + 6c = 0, which is f(2/6) * 6 = 0, or f(1/3) = 0. Since the function is continuous, if f(1/3) = 0, the root is exactly 1/3, which lies in the interval (0, 1).

AI explanation

This result is proven using the Intermediate Value Theorem. Define the polynomial f(x) = ax^2 + bx + c, and notice the condition 2a + 3b + 6c = 0. Evaluate the integral of f(x) from 0 to 1, which is (a/3) + (b/2) + c; multiplying by 6 yields 2a + 3b + 6c, meaning the area under the curve is zero. For the definite integral of a quadratic to be zero without being uniformly zero, it must cross the x-axis at least once in the interval (0, 1).

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