Multiple choice

The number or real roots of the equation $4\mathrm{x}^{3}-6\mathrm{x}^{2}+12\mathrm{x}+9=0$ is

  1. $1$
  2. $2$
  3. $3$

  4. None of these
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A Correct answer
Explanation

Let f(x) = 4x^3 - 6x^2 + 12x + 9. f'(x) = 12x^2 - 12x + 12 = 12(x^2 - x + 1). The discriminant of x^2 - x + 1 is 1 - 4 = -3, so f'(x) is always positive. The function is strictly increasing and has exactly one real root.

AI explanation

Let f(x) be the polynomial 4x^3 - 6x^2 + 12x + 9, and compute its derivative f'(x) = 12x^2 - 12x + 12. Factoring the derivative yields 12(x^2 - x + 1), and calculating the discriminant gives (-1)^2 - 4(1)(1) = -3. Because the discriminant of the derivative is negative, the derivative is always positive and the cubic function is strictly increasing everywhere. A strictly increasing polynomial crosses the x-axis exactly once, so the number of real roots is 1.

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