Question

The set \( \{ x \in \mathbb{R} : 4 + 11x - 3x^2>0 \} \) is the interval:

• A. \( \left( -\frac{1}{3}, 4 \right) \)
• B. \( \left( \frac{1}{3}, 4 \right) \)
• C. \( \left( -4, \frac{1}{3} \right) \)
• D. \( \left( -4, -\frac{1}{3} \right) \)

Hint:

For quadratic inequalities, first find the roots using the quadratic formula, then determine the intervals where the inequality holds.

Answer 1

The Correct Option is A

We are given the inequality \( 4 + 11x - 3x^2>0 \). Let's solve this quadratic inequality. First, solve the equation \( 4 + 11x - 3x^2 = 0 \). We rearrange it as: \[ 3x^2 - 11x - 4 = 0 \] Now, solve this quadratic equation using the quadratic formula: \[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \] \[ x = \frac{11 \pm \sqrt{121 + 48}}{6} = \frac{11 \pm \sqrt{169}}{6} \] \[ x = \frac{11 \pm 13}{6} \] Thus, the roots are: \[ x = \frac{11 + 13}{6} = 4 \quad \text{and} \quad x = \frac{11 - 13}{6} = -\frac{1}{3} \] The inequality \( 4 + 11x - 3x^2>0 \) holds for \( x \in \left( -\frac{1}{3}, 4 \right) \).