Solve Quadratic Inequalities

Problem: Solve the inequality

\[ 3 x^2 \gt -x + 4 \]

Solution:

1. Rewrite the inequality with one side equal to zero: \[ 3 x^2 + x - 4 \gt 0 \]
2. Find the discriminant \( \Delta \) given by: \[ \Delta = b^2 - 4ac = 1^2 - 4(3)(-4) = 49 \] Since the discriminant is positive, the left side \( 3 x^2 + x - 4 \) has two zeros at which the sign of the expression changes.
3. Factor the left side of the inequality: \[ (3x + 4)(x - 1) \gt 0 \] The two real zeros, \( -\frac{4}{3} \) and \( 1 \), divide the real number line into 3 intervals: \[ (-\infty, -\frac{4}{3}), \quad (-\frac{4}{3}, 1), \quad (1, +\infty) \]
4. Choose a real number within each interval and use it to find the sign of the expression \( (3x + 4)(x - 1) \):
   • Interval \( (-\infty, -\frac{4}{3}) \): Choose \( x = -2 \). \[ (3(-2) + 4)(-2 - 1) = (-2)(-3) = 6 \] The expression is positive.
   • Interval \( (-\frac{4}{3}, 1) \): Choose \( x = 0 \). \[ (3(0) + 4)(0 - 1) = (4)(-1) = -4 \] The expression is negative.
   • Interval \( (1, +\infty) \): Choose \( x = 4 \). \[ (3(4) + 4)(4 - 1) = (16)(3) = 48 \] The expression is positive.
5. Sign Chart: \[ \begin{array}{|c|c|c|} \hline \text{Interval} & \text{Test Point} & \text{Sign of } (3x + 4)(x - 1) \\ \hline (-\infty, -\tfrac{4}{3}) & x = -2 & + \\ \hline (-\tfrac{4}{3}, 1) & x = 0 & - \\ \hline (1, \infty) & x = 4 & + \\ \hline \end{array} \]
6. We need values of \( x \) for which the expression is greater than 0. Hence, the solution set is the union of the two intervals where the expression is positive:

   Final Answer: \[ \boxed{(-\infty, -\dfrac{4}{3}) \cup (1, +\infty)} \]

Problem: Solve the inequality

\[ -x^2 + 3x \ge -2 \]

Solution:

1. Move all terms to one side: \[ -x^2 + 3x + 2 \ge 0 \]
2. Multiply both sides by \(-1\) (flip the inequality sign): \[ x^2 - 3x - 2 \le 0 \]
3. Find the roots of the quadratic equation \( x^2 - 3x - 2 = 0 \) using the quadratic formula: \[ x = \frac{3 \pm \sqrt{17}}{2} \] Approximations:
   • \( x_1 = \frac{3 - \sqrt{17}}{2} \approx -0.56 \)
   • \( x_2 = \frac{3 + \sqrt{17}}{2} \approx 3.56 \)
4. Analyze the sign of \( x^2 - 3x - 2 \) on intervals divided by the roots: \[ \begin{array}{|c|c|c|} \hline \text{Interval} & \text{Test Point} & \text{Sign of } x^2 - 3x - 2 \\ \hline \left(-\infty, \frac{3 - \sqrt{17}}{2}\right) & x = -1 & + \\ \hline \left(\frac{3 - \sqrt{17}}{2}, \frac{3 + \sqrt{17}}{2}\right) & x = 1 & - \\ \hline \left(\frac{3 + \sqrt{17}}{2}, \infty\right) & x = 4 & + \\ \hline \end{array} \]
5. Since the inequality is \( \le 0 \), include the roots and negative interval.
   Final Answer: \[ \boxed{\left[ \frac{3 - \sqrt{17}}{2} \; , \; \frac{3 + \sqrt{17}}{2} \right]} \]

Problem: Solve the inequality

\[ x^2 \lt -x - 4 \]

Solution:

1. Rewrite the inequality with one side equal to zero: \[ x^2 + x + 4 \lt 0 \]
2. Find the discriminant \( \Delta \): \[ \Delta = b^2 - 4ac = 1^2 - 4(1)(4) = -15 \]
3. Since the discriminant is negative, the quadratic expression \( x^2 + x + 4 \) has no real zeros and keeps the same sign over the entire interval \( (-\infty, +\infty) \). To determine the sign, we can test a single value.
4. Choose \( x = 0 \) and evaluate the expression: \[ x^2 + x + 4 = 0^2 + 0 + 4 = 4 \] The expression is positive for all real \( x \), so the inequality has no solutions.