Quadratic Inequalities

Quadratic inequalities are mathematical expressions comparing the value of a quadratic function to zero, with the general form:

  • ax^2 + bx + c < 0
  • ax^2 + bx + c > 0
  • ax^2 + bx + c \le 0
  • ax^2 + bx + c \ge  0

Quadratic inequalities are similar to quadratic equations, but they involve inequality signs instead of equal signs. Solving quadratic inequalities requires finding the range of values for the variable that satisfies the inequality.

Solving Quadratic Inequalities

Solving Quadratic Inequalities by Factorisation

Factorisation is an algebraic method that can solve quadratic inequalities. To factorise quadratic inequalities, follow these steps:

Step 1: Write the quadratic inequality in the form: ax^2 + bx + c > 0 or ax^2 + bx + c < 0.

Step 2: Factor the quadratic expression on the left side of the inequality.

Step 3: Identify the critical points (the x-values that make the inequality equal to 0) and use them to form intervals.

Step 4: Test a point within each interval to determine where the inequality holds true.

Let’s look at an example:

Solve the inequality x^2 - 6x + 8 > 0 by factorisation.

1. The inequality x^2 - 6x + 8 > 0 is already in the form ax^2 + bx + c > 0.

2. Factor the quadratic expression: (x - 2)(x - 4) > 0.

3. Identify the critical points:

The critical points are x = 2 and x = 4 because these are the values of x that make the factored expression (x - 2)(x - 4) equal to 0. If you set each factor to 0, you’ll find the critical points:

x - 2 = 0x = 2

x - 4 = 0x = 4

Divide the number line into intervals using these points: x < 2, 2 < x < 4, and x > 4.

4. Test a point within each interval to determine where the inequality holds true:

a) x < 2: Choose x = 1

(1 - 2)(1 - 4) = (-1)(-3) = 3, which is greater than 0.

b) 2 < x < 4: Choose x = 3

(3 - 2)(3 - 4) = (1)(-1) = -1, which is not greater than 0.

c) x > 4: Choose x = 5

(5 - 2)(5 - 4) = (3)(1) = 3, which is greater than 0.

Based on the results of the test points, the solution set for the inequality x^2 - 6x + 8 > 0 is x < 2 or x > 4.

Solving Quadratic Inequalities Using Quadratic Formula

Using the quadratic formula is another algebraic method that can solve quadratic inequalities. To solve inequalities using the quadratic formula, follow these steps:

Step 1: Write the quadratic inequality in standard form: ax^2 + bx + c > 0 or ax^2 + bx + c < 0.

Step 2: Solve the corresponding quadratic equation ax^2 + bx + c = 0 using the quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • These solutions will be the critical points that form the intervals

Step 3: Identify the intervals where the inequality holds true by testing a point within each interval.

Let’s look at an example:

Solve the inequality 2x^2 - 5x - 3 < 0 using the quadratic formula.

1. The inequality is already in standard form: 2x^2 - 5x - 3 < 0.

2. Solve the quadratic equation 2x^2 - 5x - 3 = 0 using the quadratic formula:

x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}

x = \frac{5 \pm \sqrt{25 + 24}}{4}

x = \frac{5 \pm \sqrt{49}}{4}

x = \frac{5 \pm 7}{4}

The solutions are x = \frac{-1}{2} and x = 3. These are the critical points that divide the number line into intervals: x < \frac{-1}{2}, \frac{-1}{2} < x < 3, and x > 3.

3. Test a point within each interval to determine where the inequality holds true:

a) x < \frac{-1}{2}: Choose x = -1:

2(-1)^2 - 5(-1) - 3 = 10, which is not less than 0.

b) \frac{-1}{2} < x < 3: Choose x = 1:

2(1)^2 - 5(1) - 3 = -6, which is less than 0.

c) x > 3: Choose x = 4:

2(4)^2 - 5(4) - 3 = 21, which is not less than 0.

Based on the results of the test points, the solution set for the inequality 2x^2 - 5x - 3 < 0 is \frac{-1}{2} < x < 3.

Graphical Representation of Quadratic Inequalities

A quadratic equation of the form ax^2 + bx + c = 0 has at most two solutions. These solutions are the x-intercepts (the points at which the parabola intersects the x-axis) of the corresponding quadratic function y = ax^2 + bx + c. The quadratic equation represents the specific x-values where the function equals zero.

On the other hand, a quadratic inequality such as ax^2 + bx + c > 0 or ax^2 + bx + c < 0 has a range of x-values as its solution, rather than just two distinct points. The solution set for a quadratic inequality consists of intervals where the parabola lies above or below the x-axis, depending on the inequality sign.

When solving quadratic inequalities graphically, the graph should show the following:

  • The parabola y = ax^2 + bx + c representing the quadratic function.
  • The x-intercepts (if any) of the parabola, which are the solutions of the quadratic equation ax^2 + bx + c = 0.
  • The intervals where the parabola is above or below the x-axis, based on the inequality sign. For example, if the inequality is ax^2 + bx + c > 0, the solution set consists of the intervals where the parabola is above the x-axis.

Solving Quadratic Inequalities Graphically

To solve a quadratic inequality graphically, you can follow these steps:

Step 1: Rewrite the inequality in the form ax^2 + bx + c > 0 or ax^2 + bx + c < 0.

Step 2: Draw the corresponding quadratic function y = ax^2 + bx + c on a graph.

Step 3: Identify the x-intercepts (if any) of the parabola, which are the points where the parabola intersects the x-axis. These x-intercepts correspond to the critical points where the quadratic expression is equal to zero.

Step 4: Determine the intervals where the parabola is above or below the x-axis, based on the inequality sign.

Step 5: Combine the intervals where the inequality is true to form the solution set.

Now let’s look at two contrasting examples:

Greater than zero

If we solve the quadratic inequality x^2 - 6x + 5 > 0 graphically:

1. We draw the quadratic function y = x^2 - 6x + 5, which forms a parabola.

2. The x-intercepts are x = 1 and x = 5, which are the solutions of the quadratic equation x^2 - 6x + 5 = 0. By factoring, we get (x - 1)(x - 5) = 0.

3. The solution set for the inequality x^2 - 6x + 5 > 0 consists of the intervals where the parabola is above the x-axis. In this case, the graph shows that the parabola is above the x-axis for x < 1 and x > 5. These intervals represent the solution set for the quadratic inequality.

So, the solution set for the quadratic inequality x^2 - 6x + 5 > 0 is x < 1 or x > 5.

Less than zero

If we solve the quadratic inequality x^2 - 6x + 5 < 0 graphically:

1. We draw the quadratic function y = x^2 - 6x + 5, which forms a parabola.

2. The x-intercepts are x = 1 and x = 5, which are the solutions of the quadratic equation x^2 - 6x + 5 = 0. By factoring, we get (x - 1)(x - 5) = 0.

3. The solution set for the inequality x^2 - 6x + 5 < 0 consists of the intervals where the parabola is below the x-axis. In this case, the graph shows that the parabola is below the x-axis for 1 < x < 5. This interval represents the solution set for the quadratic inequality.

So, the solution set for the quadratic inequality x^2 - 6x + 5 < 0 is 1 < x < 5.

Greater than zero vs less than zero

Comparing the two examples:

  • For the inequality x^2 - 6x + 5 > 0, the solution set is x < 1 or x > 5. The graph shows the parabola above the x-axis in these intervals.
  • For the inequality x^2 - 6x + 5 < 0, the solution set is 1 < x < 5. The graph shows the parabola below the x-axis in this interval.

Examples

Example 1:

Solve the inequality x^2 - 4x + 3 > 0.

1. First, factor the quadratic expression: (x - 1)(x - 3) > 0(x - 1)(x - 3) > 0.

2. Identify the critical points: x = 1x = 1 and x = 3x = 3.

3. Test the intervals between the critical points to find where the inequality holds true. In this case, test x < 1x < 1, 1 < x < 31 < x < 3, and x > 3x > 3.

4. The inequality is true when x < 1x < 1 and x > 3x > 3.

The solution set for this inequality is x < 1x < 1 or x > 3x > 3.

Example 2:

Solve the inequality x^2 - 6x + 8 \le 0.

1. Factor the quadratic expression: (x - 2)(x - 4) \leq 0(x - 2)(x - 4) \leq 0.

2. Identify the critical points: x = 2x = 2 and x = 4x = 4.

3. Test the intervals between the critical points: x < 2, 2 \leq x \leq 4x < 2, 2 \leq x \leq 4, and x > 4x > 4.

4. The inequality is true when 2 \leq x \leq 42 \leq x \leq 4.

The solution set for this inequality is 2 \leq x \leq 42 \leq x \leq 4.

Example 3:

A rectangular garden has a width of x meters and a length of 2x meters. The garden’s area must be less than or equal to 60 square meters. Determine the possible values of x.

1. Write the inequality: x(2x) \leq 60x(2x) \leq 60.

2. Simplify the inequality: 2x^2 \leq 602x^2 \leq 60.

3. Divide both sides by 2: x^2 \leq 30x^2 \leq 30.

4. To find the possible values of x, consider the square root of both sides: -\sqrt{30} \leq x \leq \sqrt{30}-\sqrt{30} \leq x \leq \sqrt{30}.

The solution set for this inequality is -\sqrt{30} \leq x \leq \sqrt{30}-\sqrt{30} \leq x \leq \sqrt{30}. However, since width cannot be negative, the practical solution set for this problem is 0 < x \leq \sqrt{30}0 < x \leq \sqrt{30}.

Example 4:

Solve the quadratic inequality x^2 - 4x + 3 > 0 graphically.

1. The inequality is already in standard form: x^2 - 4x + 3 > 0x^2 - 4x + 3 > 0

2. Draw the corresponding quadratic function y = x^2 - 4x + 3y = x^2 - 4x + 3 on a graph, which will form a parabola.

 

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3. Identify the x-intercepts of the parabola. In this case, the parabola intersects the x-axis at x = 1x = 1 and x = 3x = 3. These are the critical points.

4. Determine the intervals where the parabola is above the x-axis, as the inequality is greater than zero (x^2 - 4x + 3 > 0)(x^2 - 4x + 3 > 0). Based on the graph, the parabola is above the x-axis for x < 1x < 1 and x > 3x > 3.

5. The solution set for the inequality x^2 - 4x + 3 > 0x^2 - 4x + 3 > 0 is x < 1x < 1 or x > 3x > 3.