Solve Quadratic Inequalities

This tutorial explains how to solve quadratic inequalities of the form \( ax^2 + bx + c \ge 0 \) and \( ax^2 + bx + c \le 0 \), using clear examples and detailed solutions. The method involves finding the zeros (or roots) of the corresponding quadratic equation, factoring or using the quadratic formula, and then analyzing the sign of the expression over different intervals. Solutions are presented in interval notation, supported by a sign chart or number line to help visualize where the expression is positive, negative, or zero.

Example 1:

Solve the inequality \[ 3 x^2 \gt -x + 4 \]

Solution to Example 1:

Given \[ 3 x^2 > -x + 4 \] Rewrite the inequality with one side equal to zero. \[ 3 x^2 + x - 4 > 0 \] Find the discriminant \( \Delta \) given by \[ \Delta = b^2 - 4 a c = 1^2 - 4 (3) (-4) = 49 \] Since the discriminant is positive, the left side \( 3 x^2 + x - 4 \) of the inequality has two zeros at which the sign of the expression \( 3 x^2 + x - 4 \) changes.

Factor the left side of the inequality. \[ (3x + 4)(x - 1) \gt 0 \] The two real zeros \( - 4 / 3 \) and \( 1 \) of the left side of the inequality, divide the real number line into 3 intervals. \[ (-\infty, -\dfrac{4}{3}) , \quad (-\dfrac{4}{3}, 1) \quad , (1, +\infty) \] We chose a real number within each interval and use it to find the sign of the expression \( (3x + 4)(x - 1) \).

a) interval \( (-\infty, -\dfrac{4}{3}) \) : chose a value of x within the interval, for example \( x = - 2 \) and find the sign of \( (3x + 4)(x - 1) \)
\[ (3x + 4)(x - 1) = (3(-2) + 4)(-2 - 1) = 6 \] Hence The expression \( (3x + 4)(x - 1) \) is positive in the interval \( (-\infty, -\dfrac{4}{3}) \)
b) interval \( (-\dfrac{4}{3}, 1) \) : chose a value of x within the interval, for example \( x = 0 \) and find the sign of \[ (3x + 4)(x - 1) = (0 + 4)(0 - 1) = - 4 \] Hence the expression \( (3x + 4)(x - 1) \) is negative in the inteval \( (-\dfrac{4}{3}, 1) \)
c) interval \( (1, +\infty) \) : chose a value of x within the interval, for example \( x = 4 \) and find the sign of \[ (3x + 4)(x - 1) = (3(4) + 4)((4) - 1) = 48 \] Hence the expression \( (3x + 4)(x - 1) \) is positive in the interval \( (1, +\infty) \)

Sign chart \[ \begin{array}{|c|c|c|} \hline \text{Interval} & \text{Test Point} & \text{Sign of } (3x + 4)(x - 1) \\ \hline (-\infty, -\tfrac{4}{3}) & x = -2 & + \\ \hline (-\tfrac{4}{3}, 1) & x = 0 & - \\ \hline (1, \infty) & x = 4 & + \\ \hline \end{array} \] We need values of \( x \) for which the expressions \( (3x + 4)(x - 1) \) is greater than 0, hence the solution set is the union of the two interval for which \( (3x + 4)(x - 1) \) is greater than 0 and is given by: \[ \boxed{ (-\infty, -\dfrac{4}{3}) \cup (1, +\infty) } \]

Example 2

Solve the inequality \[ -x^2 + 3x \ge -2 \]

Solution to Example 2:

Move all terms to one side:

\[ -x^2 + 3x + 2 \ge 0 \]

Multiply both sides by \(-1\) (flip the inequality sign):

\[ x^2 - 3x - 2 \le 0 \]

Find the roots of the quadratic equation \( x^2 - 3x - 2 = 0 \) using the quadratic formula:

\[ x = \frac{3 \pm \sqrt{17}}{2} \]

Approximations:

\( x_1 = \frac{3 - \sqrt{17}}{2} \approx -0.56 \)
\( x_2 = \frac{3 + \sqrt{17}}{2} \approx 3.56 \)

Analyze the sign of \( x^2 - 3x - 2 \) on intervals divided by the roots:

\[ \begin{array}{|c|c|c|} \hline \text{Interval} & \text{Test Point} & \text{Sign of } x^2 - 3x - 2 \\ \hline \left(-\infty, \frac{3 - \sqrt{17}}{2}\right) & x = -1 & + \\ \hline \left(\frac{3 - \sqrt{17}}{2}, \frac{3 + \sqrt{17}}{2}\right) & x = 1 & - \\ \hline \left(\frac{3 + \sqrt{17}}{2}, \infty\right) & x = 4 & + \\ \hline \end{array} \]

Since the inequality is \( \le 0 \), include the roots and negative interval. The solution is:

\[ \boxed{ \left[ \frac{3 - \sqrt{17}}{2} \; , \; \frac{3 + \sqrt{17}}{2} \right] } \]

Example 3

Solve the inequality \[ x^2 \lt -x - 4 \]

Solution to Example 3:

Given: \[ x^2 < -x - 4 \]

Rewrite the inequality with one side equal to zero:

\[ x^2 + x + 4 < 0 \]

Find the discriminant \( \Delta \):

\[ \Delta = b^2 - 4ac = 1^2 - 4(1)(4) = -15 \]

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.

Choose \(x = 0\) and evaluate the expression:

\[ x^2 + x + 4 = 0^2 + 0 + 4 = 4 \]

The expression \(x^2 + x + 4\) is positive for all real \(x\), so the inequality has no solutions.

Exercises

Solve the following quadratic inequalities and express your answers using interval notation.

\( -x^2 + 2x > -3 \)
\( x^2 - 4x > -6 \)

Solutions to the Above Exercises

\( (-1, 3) \)
\( (-\infty, +\infty) \)

More references and links on how to Solve Equations, Systems of Equations and Inequalities Step by Step Solver for Quadratic Inequalities