Solving Inequalities A Comprehensive Guide With Examples

Pleton · · 5 min read

Table Of Content

  1. Solving Quadratic Inequalities: A Step-by-Step Approach
    1. Example 1: Solving (x - 5)(x + 3) ≥ 0
    2. Example 2: Solving (x - 3)(x - 5) < 0
  2. Tackling More Quadratic Inequalities
    1. Example 3: Solving (x + 4)(x - 7) > 0
    2. Example 4: Solving x² + 10x + 21 ≥ 0
    3. Example 5: Solving x² + 6x + 5 > 0
    4. Example 6: Solving x² - 3x + 2 < 0
  3. Additional Practice Problems
    1. Example 7: Solving x² + 7x + 12 ≥ 0
    2. Example 8: Solving x² - 12x + 35 ≤ 0
    3. Example 9: Solving x² + x - 6 ≤ 0
  4. Mastering Inequalities: Your Path to Success

Hey everyone! Inequalities might seem tricky at first, but trust me, with a bit of practice, you'll be solving them like a pro. This guide will walk you through various types of inequalities, from basic quadratics to more complex scenarios. We'll break down each step, making sure you understand the why behind the how. So, let's dive in and conquer those inequalities!

Solving Quadratic Inequalities: A Step-by-Step Approach

When tackling quadratic inequalities, the key is to understand the behavior of quadratic functions. Remember, a quadratic function forms a parabola, and the solutions to the inequality depend on whether the parabola opens upwards or downwards and where it intersects the x-axis. Let's start with the fundamental steps involved in solving quadratic inequalities, and then we'll apply these steps to specific examples.

  1. Rewrite the Inequality: The first step is to ensure the inequality is in the standard form, which is either ax² + bx + c ≥ 0, ax² + bx + c ≤ 0, ax² + bx + c > 0, or ax² + bx + c < 0. This standard form makes it easier to identify the coefficients and constants involved. For instance, if you have an inequality like x² + 10x ≥ -21, you need to rewrite it as x² + 10x + 21 ≥ 0 by adding 21 to both sides. Similarly, if you encounter an inequality like 2x² < 5x + 3, you should transform it into 2x² - 5x - 3 < 0 by subtracting 5x and 3 from both sides. Getting the inequality into this standard form is crucial because it sets the stage for the next steps, such as factoring and finding critical points. Without this initial step, it becomes significantly harder to proceed with solving the inequality accurately.

  2. Factor the Quadratic Expression: Once you have the inequality in standard form, the next step is to factor the quadratic expression, that is, the ax² + bx + c part. Factoring the quadratic expression allows us to find the roots or zeros of the quadratic equation, which are the x-values where the parabola intersects the x-axis. These roots are essential because they divide the number line into intervals, and the sign of the quadratic expression remains constant within each interval. For example, if you have the inequality x² + 5x + 6 > 0, you would factor the quadratic expression x² + 5x + 6 into (x + 2)(x + 3). Similarly, for the inequality x² - 4x + 3 < 0, you would factor x² - 4x + 3 into (x - 1)(x - 3). Factoring can be done using various techniques, such as trial and error, the quadratic formula, or completing the square. The goal is to express the quadratic expression as a product of two linear factors, which then makes it easier to identify the critical points.

  3. Identify the Critical Points: The critical points are the values of x that make the factored expression equal to zero. These points are crucial because they divide the number line into intervals where the quadratic expression is either positive or negative. To find the critical points, set each factor equal to zero and solve for x. For example, if your factored expression is (x + 2)(x - 3), you would set (x + 2) = 0 and (x - 3) = 0. Solving these equations gives you the critical points x = -2 and x = 3. These critical points are the x-intercepts of the parabola represented by the quadratic function. They are the points where the graph crosses the x-axis, and they play a vital role in determining the intervals where the inequality holds true. Identifying the critical points is a fundamental step in solving quadratic inequalities because it provides the boundaries for testing intervals.

  4. Create a Sign Chart: A sign chart is a visual tool that helps determine the sign of the quadratic expression in each interval created by the critical points. Draw a number line and mark the critical points on it. These points divide the number line into several intervals. For each interval, choose a test value (a number within the interval) and plug it into the factored form of the quadratic expression. The sign of the result will tell you whether the expression is positive or negative in that interval. For instance, if your critical points are -2 and 3, you'll have three intervals to consider: (-∞, -2), (-2, 3), and (3, ∞). Pick a test value from each interval, such as -3, 0, and 4, respectively. Plug these values into the factored expression and note the sign of the result. This sign chart provides a clear picture of how the quadratic expression behaves across the number line, making it easier to identify the intervals that satisfy the inequality.

  5. Determine the Solution Intervals: Based on the sign chart and the original inequality, identify the intervals where the inequality is satisfied. If the inequality is of the form ax² + bx + c > 0 or ax² + bx + c ≥ 0, you're looking for intervals where the expression is positive or positive and zero. Conversely, if the inequality is of the form ax² + bx + c < 0 or ax² + bx + c ≤ 0, you're looking for intervals where the expression is negative or negative and zero. It's crucial to pay attention to whether the inequality includes an "equal to" sign (≥ or ≤) because, in that case, the critical points themselves are also part of the solution. For example, if you're solving x² - 4x + 3 < 0 and your sign chart shows that the expression is negative between the critical points 1 and 3, the solution is the interval (1, 3). However, if the inequality is x² - 4x + 3 ≤ 0, the solution would include the critical points, making it the closed interval [1, 3]. This step is where you translate the information from the sign chart into the final solution set.

  6. Write the Solution: Finally, express the solution in interval notation. Interval notation is a concise way to represent the set of all values that satisfy the inequality. Use parentheses ( or ) to indicate that the endpoint is not included in the solution (for strict inequalities < or >) and square brackets [ or ] to indicate that the endpoint is included (for inequalities with ≤ or ≥). If the solution consists of multiple intervals, use the union symbol ∪ to combine them. For example, if the solution to an inequality is all real numbers less than -2 or greater than 3, you would write this in interval notation as (-∞, -2) ∪ (3, ∞). If the solution is all real numbers between -1 and 4, including -1 and 4, you would write [-1, 4]. Correctly expressing the solution in interval notation is the final step in communicating the answer in a standard mathematical format.

Example 1: Solving (x - 5)(x + 3) ≥ 0

Let's tackle this inequality step-by-step:

  1. Rewrite: The inequality is already in a factored form, which is super convenient!
  2. Factor: It's already factored as (x - 5)(x + 3) ≥ 0.
  3. Critical Points: Set each factor to zero:
    • x - 5 = 0 => x = 5
    • x + 3 = 0 => x = -3
  4. Sign Chart:
    • Create a number line and mark -3 and 5.
    • Test intervals: (-∞, -3), (-3, 5), (5, ∞).
      • Test x = -4: (-4 - 5)(-4 + 3) = (-9)(-1) = 9 > 0
      • Test x = 0: (0 - 5)(0 + 3) = (-5)(3) = -15 < 0
      • Test x = 6: (6 - 5)(6 + 3) = (1)(9) = 9 > 0
  5. Solution Intervals: We want where the expression is greater than or equal to 0, so we consider the intervals where the expression is positive or zero. This gives us (-∞, -3] and [5, ∞).
  6. Solution: The solution in interval notation is (-∞, -3] ∪ [5, ∞).

Example 2: Solving (x - 3)(x - 5) < 0

Let's walk through another example:

  1. Rewrite: The inequality is already in factored form.
  2. Factor: Already factored as (x - 3)(x - 5) < 0.
  3. Critical Points: Set each factor to zero:
    • x - 3 = 0 => x = 3
    • x - 5 = 0 => x = 5
  4. Sign Chart:
    • Create a number line and mark 3 and 5.
    • Test intervals: (-∞, 3), (3, 5), (5, ∞).
      • Test x = 2: (2 - 3)(2 - 5) = (-1)(-3) = 3 > 0
      • Test x = 4: (4 - 3)(4 - 5) = (1)(-1) = -1 < 0
      • Test x = 6: (6 - 3)(6 - 5) = (3)(1) = 3 > 0
  5. Solution Intervals: We want where the expression is less than 0, so we consider the interval where the expression is negative. This gives us (3, 5).
  6. Solution: The solution in interval notation is (3, 5).

Tackling More Quadratic Inequalities

Let's reinforce our understanding with a few more examples, each presenting a slightly different scenario. By working through these examples, you'll become more comfortable applying the steps we've discussed and gain confidence in solving a wider range of quadratic inequalities.

Example 3: Solving (x + 4)(x - 7) > 0

This inequality is already factored, making our job a bit easier. Let's break it down:

  1. Rewrite: The inequality is already in factored form.
  2. Factor: Factored as (x + 4)(x - 7) > 0.
  3. Critical Points: Set each factor to zero:
    • x + 4 = 0 => x = -4
    • x - 7 = 0 => x = 7
  4. Sign Chart:
    • Create a number line and mark -4 and 7.
    • Test intervals: (-∞, -4), (-4, 7), (7, ∞).
      • Test x = -5: (-5 + 4)(-5 - 7) = (-1)(-12) = 12 > 0
      • Test x = 0: (0 + 4)(0 - 7) = (4)(-7) = -28 < 0
      • Test x = 8: (8 + 4)(8 - 7) = (12)(1) = 12 > 0
  5. Solution Intervals: We want where the expression is greater than 0, so we consider the intervals where the expression is positive. This gives us (-∞, -4) and (7, ∞).
  6. Solution: The solution in interval notation is (-∞, -4) ∪ (7, ∞).

Example 4: Solving x² + 10x + 21 ≥ 0

Now, let's tackle one where we need to factor the quadratic expression ourselves:

  1. Rewrite: The inequality is already in standard form.
  2. Factor: Factor x² + 10x + 21 into (x + 3)(x + 7).
  3. Critical Points: Set each factor to zero:
    • x + 3 = 0 => x = -3
    • x + 7 = 0 => x = -7
  4. Sign Chart:
    • Create a number line and mark -7 and -3.
    • Test intervals: (-∞, -7), (-7, -3), (-3, ∞).
      • Test x = -8: (-8 + 3)(-8 + 7) = (-5)(-1) = 5 > 0
      • Test x = -4: (-4 + 3)(-4 + 7) = (-1)(3) = -3 < 0
      • Test x = 0: (0 + 3)(0 + 7) = (3)(7) = 21 > 0
  5. Solution Intervals: We want where the expression is greater than or equal to 0, so we consider the intervals where the expression is positive or zero. This gives us (-∞, -7] and [-3, ∞).
  6. Solution: The solution in interval notation is (-∞, -7] ∪ [-3, ∞).

Example 5: Solving x² + 6x + 5 > 0

Let's continue practicing our factoring skills:

  1. Rewrite: The inequality is already in standard form.
  2. Factor: Factor x² + 6x + 5 into (x + 1)(x + 5).
  3. Critical Points: Set each factor to zero:
    • x + 1 = 0 => x = -1
    • x + 5 = 0 => x = -5
  4. Sign Chart:
    • Create a number line and mark -5 and -1.
    • Test intervals: (-∞, -5), (-5, -1), (-1, ∞).
      • Test x = -6: (-6 + 1)(-6 + 5) = (-5)(-1) = 5 > 0
      • Test x = -2: (-2 + 1)(-2 + 5) = (-1)(3) = -3 < 0
      • Test x = 0: (0 + 1)(0 + 5) = (1)(5) = 5 > 0
  5. Solution Intervals: We want where the expression is greater than 0, so we consider the intervals where the expression is positive. This gives us (-∞, -5) and (-1, ∞).
  6. Solution: The solution in interval notation is (-∞, -5) ∪ (-1, ∞).

Example 6: Solving x² - 3x + 2 < 0

One more example to solidify our understanding:

  1. Rewrite: The inequality is already in standard form.
  2. Factor: Factor x² - 3x + 2 into (x - 1)(x - 2).
  3. Critical Points: Set each factor to zero:
    • x - 1 = 0 => x = 1
    • x - 2 = 0 => x = 2
  4. Sign Chart:
    • Create a number line and mark 1 and 2.
    • Test intervals: (-∞, 1), (1, 2), (2, ∞).
      • Test x = 0: (0 - 1)(0 - 2) = (-1)(-2) = 2 > 0
      • Test x = 1.5: (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0
      • Test x = 3: (3 - 1)(3 - 2) = (2)(1) = 2 > 0
  5. Solution Intervals: We want where the expression is less than 0, so we consider the interval where the expression is negative. This gives us (1, 2).
  6. Solution: The solution in interval notation is (1, 2).

Additional Practice Problems

To further hone your skills, let's tackle some more inequalities. These examples will help you become even more comfortable with the process and develop a deeper understanding of how different quadratic expressions behave. Remember, the key is to practice consistently and apply the steps we've discussed. So, grab a pencil and paper, and let's work through these problems together!

Example 7: Solving x² + 7x + 12 ≥ 0

Let's get started with this one. Follow the same steps we've been using:

  1. Rewrite: The inequality is already in the standard form.
  2. Factor: We need to factor x² + 7x + 12. Think of two numbers that multiply to 12 and add up to 7. These numbers are 3 and 4. So, the factored form is (x + 3)(x + 4) ≥ 0.
  3. Critical Points: Set each factor equal to zero:
    • x + 3 = 0 => x = -3
    • x + 4 = 0 => x = -4
  4. Sign Chart:
    • Draw a number line and mark the critical points -4 and -3.
    • This divides the number line into three intervals: (-∞, -4), (-4, -3), and (-3, ∞).
    • Pick test values in each interval:
      • For (-∞, -4), let's use x = -5: (-5 + 3)(-5 + 4) = (-2)(-1) = 2 > 0
      • For (-4, -3), let's use x = -3.5: (-3.5 + 3)(-3.5 + 4) = (-0.5)(0.5) = -0.25 < 0
      • For (-3, ∞), let's use x = 0: (0 + 3)(0 + 4) = (3)(4) = 12 > 0
  5. Solution Intervals: We want the intervals where the expression is greater than or equal to 0. From the sign chart, these are (-∞, -4] and [-3, ∞). We include the critical points because the inequality includes "or equal to".
  6. Solution: The solution in interval notation is (-∞, -4] ∪ [-3, ∞).

Example 8: Solving x² - 12x + 35 ≤ 0

Let's try another one. This time, we're looking for values less than or equal to zero.

  1. Rewrite: The inequality is already in standard form.
  2. Factor: We need to factor x² - 12x + 35. We're looking for two numbers that multiply to 35 and add up to -12. These numbers are -5 and -7. So, the factored form is (x - 5)(x - 7) ≤ 0.
  3. Critical Points: Set each factor equal to zero:
    • x - 5 = 0 => x = 5
    • x - 7 = 0 => x = 7
  4. Sign Chart:
    • Draw a number line and mark the critical points 5 and 7.
    • This divides the number line into three intervals: (-∞, 5), (5, 7), and (7, ∞).
    • Pick test values in each interval:
      • For (-∞, 5), let's use x = 4: (4 - 5)(4 - 7) = (-1)(-3) = 3 > 0
      • For (5, 7), let's use x = 6: (6 - 5)(6 - 7) = (1)(-1) = -1 < 0
      • For (7, ∞), let's use x = 8: (8 - 5)(8 - 7) = (3)(1) = 3 > 0
  5. Solution Intervals: We want the intervals where the expression is less than or equal to 0. From the sign chart, this is the interval [5, 7]. We include the critical points because of the "or equal to".
  6. Solution: The solution in interval notation is [5, 7].

Example 9: Solving x² + x - 6 ≤ 0

Here's another example to keep practicing those factoring skills.

  1. Rewrite: The inequality is already in standard form.
  2. Factor: We need to factor x² + x - 6. We're looking for two numbers that multiply to -6 and add up to 1. These numbers are 3 and -2. So, the factored form is (x + 3)(x - 2) ≤ 0.
  3. Critical Points: Set each factor equal to zero:
    • x + 3 = 0 => x = -3
    • x - 2 = 0 => x = 2
  4. Sign Chart:
    • Draw a number line and mark the critical points -3 and 2.
    • This divides the number line into three intervals: (-∞, -3), (-3, 2), and (2, ∞).
    • Pick test values in each interval:
      • For (-∞, -3), let's use x = -4: (-4 + 3)(-4 - 2) = (-1)(-6) = 6 > 0
      • For (-3, 2), let's use x = 0: (0 + 3)(0 - 2) = (3)(-2) = -6 < 0
      • For (2, ∞), let's use x = 3: (3 + 3)(3 - 2) = (6)(1) = 6 > 0
  5. Solution Intervals: We want the intervals where the expression is less than or equal to 0. From the sign chart, this is the interval [-3, 2]. We include the critical points because of the "or equal to".
  6. Solution: The solution in interval notation is [-3, 2].

Mastering Inequalities: Your Path to Success

Alright, guys, you've made it through a comprehensive guide to solving inequalities! We've covered everything from the basic steps to tackling various quadratic inequalities. Remember, the key to mastering inequalities is practice. Work through these examples, try additional problems, and don't be afraid to make mistakes – they're part of the learning process. With consistent effort, you'll become confident in your ability to solve any inequality that comes your way.

So, keep practicing, and remember, you've got this! Inequalities might have seemed daunting at first, but now you have the tools and knowledge to conquer them. Go out there and solve those inequalities like a pro!

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