Quadratic Inequalities

A quadratic inequality is an inequality that can be written in the form:

$$ ax^2 + bx + c > 0, $$

$$ ax^2 + bx + c < 0, $$

$$ ax^2 + bx + c \geq 0, $$

$$ ax^2 + bx + c \leq 0, $$

where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. Solving a quadratic inequality involves finding the range of values of \( x \) that make the inequality true.

How to Solve Quadratic Inequalities

  1. Rewrite the inequality in standard form (if necessary) as \( ax^2 + bx + c \; \{>,<,\geq,\leq\} \; 0 \).
  2. Find the roots of the corresponding quadratic equation \( ax^2 + bx + c = 0 \) by factoring, completing the square, or using the quadratic formula.
  3. Use the roots to divide the number line into intervals.
  4. Test a value from each interval in the inequality to determine which intervals satisfy it.

Example: Solving a Quadratic Inequality

Solve the inequality:

$$ x^2 - 5x + 6 > 0 $$

Step 1: Find the roots of the equation \( x^2 - 5x + 6 = 0 \):

$$ (x - 2)(x - 3) = 0 $$

The roots are \( x = 2 \) and \( x = 3 \).

Step 2: Divide the number line into intervals based on the roots: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).

Step 3: Test a value from each interval in the inequality:

  • For \( x = 1 \) in \( (-\infty, 2) \): \( (1)^2 - 5(1) + 6 = 2 \) (True).
  • For \( x = 2.5 \) in \( (2, 3) \): \( (2.5)^2 - 5(2.5) + 6 = -0.25 \) (False).
  • For \( x = 4 \) in \( (3, \infty) \): \( (4)^2 - 5(4) + 6 = 2 \) (True).

Thus, the solution is \( x < 2 \) or \( x > 3 \).

Practice Questions

  1. Solve the inequality \( x^2 + 4x - 5 \geq 0 \).
    Solution

    Rewrite as \( x^2 + 4x - 5 = 0 \):

    $$ (x + 5)(x - 1) = 0 $$

    The roots are \( x = -5 \) and \( x = 1 \).

    Test intervals \( (-\infty, -5) \), \( (-5, 1) \), \( (1, \infty) \).

    • For \( x = -6 \): \( (-6)^2 + 4(-6) - 5 = 7 \) (True).
    • For \( x = 0 \): \( (0)^2 + 4(0) - 5 = -5 \) (False).
    • For \( x = 2 \): \( (2)^2 + 4(2) - 5 = 3 \) (True).

    Solution: \( x \leq -5 \) or \( x \geq 1 \).

  2. Solve the inequality \( 2x^2 - 3x \leq 5 \).
    Solution

    Rewrite as \( 2x^2 - 3x - 5 \leq 0 \):

    Using the quadratic formula to find the roots:

    $$ x = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} $$

    • Root 1: \( x = \frac{10}{4} = 2.5 \)
    • Root 2: \( x = \frac{-4}{4} = -1 \)

    Test intervals \( (-\infty, -1) \), \( (-1, 2.5) \), \( (2.5, \infty) \).

    • For \( x = -2 \): \( 2(-2)^2 - 3(-2) - 5 = 3 \) (False).
    • For \( x = 0 \): \( 2(0)^2 - 3(0) - 5 = -5 \) (True).
    • For \( x = 3 \): \( 2(3)^2 - 3(3) - 5 = 4 \) (False).

    Solution: \( -1 \leq x \leq 2.5 \).

  3. Solve the inequality \( x^2 - 2x < 8 \).
    Solution

    Rewrite as \( x^2 - 2x - 8 < 0 \):

    Factor as:

    $$ (x - 4)(x + 2) = 0 $$

    The roots are \( x = 4 \) and \( x = -2 \).

    Test intervals \( (-\infty, -2) \), \( (-2, 4) \), \( (4, \infty) \).

    • For \( x = -3 \): \( (-3)^2 - 2(-3) - 8 = 1 \) (False).
    • For \( x = 0 \): \( (0)^2 - 2(0) - 8 = -8 \) (True).
    • For \( x = 5 \): \( (5)^2 - 2(5) - 8 = 7 \) (False).

    Solution: \( -2 < x < 4 \).

→Applications of Quadratics in Real Life