Solving Quadratic Inequalities: A Step-by-Step Guide

by Alex Johnson 53 views

Let's dive into the world of quadratic inequalities! If you've ever stumbled upon an inequality like 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0 and felt a bit lost, you're in the right place. This guide will break down the process step-by-step, making it easy to understand and implement. We'll cover everything from factoring quadratic expressions to interpreting the results on a number line. So, grab a pen and paper, and let's get started!

Understanding Quadratic Inequalities

Before we jump into solving, let's make sure we're on the same page about what a quadratic inequality is. A quadratic inequality is an inequality that involves a quadratic expression. A quadratic expression is a polynomial of degree two, generally in the form ax2+bx+cax^2 + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. When we set this expression to be greater than, less than, greater than or equal to, or less than or equal to another value (often zero), we get a quadratic inequality.

For example, 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0 is a quadratic inequality. The goal is to find the range(s) of values for 'x' that make the inequality true. This is different from solving a quadratic equation (where we look for specific values of 'x' that make the expression equal to zero), as we're now looking for intervals of 'x' values.

Why are Quadratic Inequalities Important?

You might be wondering, "Why should I bother learning about quadratic inequalities?" Well, they pop up in various real-world applications, from physics to economics. For instance, they can be used to model projectile motion, optimize business profits, or determine the stability of systems. Understanding how to solve them is a valuable skill in many fields.

Furthermore, the techniques we learn for solving quadratic inequalities also apply to more complex inequalities and are fundamental to understanding calculus and other advanced mathematical topics. So, mastering this concept is a stepping stone to more advanced mathematical problem-solving.

Step 1: Rewrite the Inequality (if Necessary)

The first step in solving a quadratic inequality is to make sure it's in the standard form: ax2+bx+c>0ax^2 + bx + c > 0, ax2+bx+c<0ax^2 + bx + c < 0, ax2+bx+c">="0ax^2 + bx + c ">= "0, or ax2+bx+c"<="0ax^2 + bx + c "<=" 0. This means you might need to rearrange the terms so that all the terms are on one side of the inequality, and zero is on the other side.

In our example, 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0, the inequality is already in the standard form, so we can move on to the next step. However, if we had an inequality like 2x2>9xβˆ’42x^2 > 9x - 4, we would need to subtract 9x9x and add 44 to both sides to get it into the standard form.

Why is this step important? Having the inequality in standard form allows us to easily identify the coefficients 'a', 'b', and 'c', which we'll need for the next steps. It also sets the stage for using methods like factoring or the quadratic formula effectively.

Step 2: Find the Roots of the Quadratic Equation

The next crucial step is to find the roots (or solutions) of the corresponding quadratic equation. That is, we replace the inequality sign with an equals sign and solve the equation ax2+bx+c=0ax^2 + bx + c = 0. These roots are the points where the quadratic expression equals zero, and they are the critical points that divide the number line into intervals.

For our example, 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0, we need to solve the equation 2x2βˆ’9x+4=02x^2 - 9x + 4 = 0. There are several ways to do this:

  1. Factoring: We look for two numbers that multiply to give (2)(4)=8(2)(4) = 8 and add up to βˆ’9-9. These numbers are βˆ’8-8 and βˆ’1-1. So, we can rewrite the middle term:

    2x2βˆ’8xβˆ’x+4=02x^2 - 8x - x + 4 = 0

    Now, we factor by grouping:

    2x(xβˆ’4)βˆ’1(xβˆ’4)=02x(x - 4) - 1(x - 4) = 0

    (2xβˆ’1)(xβˆ’4)=0(2x - 1)(x - 4) = 0

    This gives us two roots: x = rac{1}{2} and x=4x = 4.

  2. Quadratic Formula: If factoring is difficult, we can use the quadratic formula:

    x = rac{-b "Β±" \sqrt{b^2 - 4ac}}{2a}

    For our equation, a=2a = 2, b=βˆ’9b = -9, and c=4c = 4. Plugging these values in:

    x = rac{-(-9) "Β±" \sqrt{(-9)^2 - 4(2)(4)}}{2(2)}

    x = rac{9 "Β±" \sqrt{81 - 32}}{4}

    x = rac{9 "Β±" \sqrt{49}}{4}

    x = rac{9 "Β±" 7}{4}

    This gives us the same roots: x = rac{1}{2} and x=4x = 4.

Why find the roots? The roots are crucial because they are the points where the parabola (the graph of the quadratic function) intersects the x-axis. These points divide the x-axis into intervals where the quadratic expression is either positive or negative.

Step 3: Create a Sign Chart

Now that we have the roots, we can create a sign chart. A sign chart is a visual tool that helps us determine the intervals where the quadratic expression is positive or negative. It's a number line with the roots marked on it, dividing the line into intervals.

  1. Draw a number line: Draw a horizontal line and mark the roots we found in Step 2 on the line. In our example, we mark 12\frac{1}{2} and 44 on the number line. These points divide the number line into three intervals: (βˆ’βˆž,12)(-\infty, \frac{1}{2}), (12,4)(\frac{1}{2}, 4), and (4,∞)(4, \infty).

  2. Choose test values: Pick a test value within each interval. For example:

    • In the interval (βˆ’βˆž,12)(-\infty, \frac{1}{2}), we can choose x=0x = 0.
    • In the interval (12,4)(\frac{1}{2}, 4), we can choose x=1x = 1.
    • In the interval (4,∞)(4, \infty), we can choose x=5x = 5.
  3. Evaluate the quadratic expression: Plug each test value into the quadratic expression 2x2βˆ’9x+42x^2 - 9x + 4 and determine the sign (positive or negative) of the result:

    • For x=0x = 0: 2(0)2βˆ’9(0)+4=42(0)^2 - 9(0) + 4 = 4 (positive)
    • For x=1x = 1: 2(1)2βˆ’9(1)+4=2βˆ’9+4=βˆ’32(1)^2 - 9(1) + 4 = 2 - 9 + 4 = -3 (negative)
    • For x=5x = 5: 2(5)2βˆ’9(5)+4=50βˆ’45+4=92(5)^2 - 9(5) + 4 = 50 - 45 + 4 = 9 (positive)
  4. Mark the signs on the sign chart: Write the sign of the result above each interval on the number line. In our case, we have:

    • (βˆ’βˆž,12)(-\infty, \frac{1}{2}): +
    • (12,4)(\frac{1}{2}, 4): -
    • (4,∞)(4, \infty): +

Why use a sign chart? The sign chart provides a clear visual representation of where the quadratic expression is positive, negative, or zero. This makes it easy to identify the solution intervals for the inequality.

Step 4: Determine the Solution

Now we come to the final step: determining the solution to the inequality. We use the sign chart we created in Step 3 to find the intervals that satisfy the original inequality.

Our original inequality was 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0. This means we're looking for the intervals where the quadratic expression is greater than zero, i.e., positive.

Looking at our sign chart, we see that the expression is positive in the intervals (βˆ’βˆž,12)(-\infty, \frac{1}{2}) and (4,∞)(4, \infty). Therefore, these are the solution intervals.

Since the inequality is strictly greater than zero (>>), we don't include the roots themselves in the solution. If the inequality were greater than or equal to zero (>=>=), we would include the roots.

So, the solution to the inequality 2x2βˆ’9x+4>02x^2 - 9x + 4 > 0 is:

x<12x < \frac{1}{2} or x>4x > 4

In interval notation, this is written as:

(βˆ’βˆž,12)βˆͺ(4,∞)(-\infty, \frac{1}{2}) \cup (4, \infty)

Why is this the final step? This step directly answers the question posed by the inequality. By using the sign chart, we've identified the ranges of 'x' values that make the inequality true.

Alternative Scenarios

Let's briefly consider how the solution would change if the inequality were different:

  1. If the inequality were 2x2βˆ’9x+4<02x^2 - 9x + 4 < 0, we would look for the intervals where the expression is negative. From our sign chart, this is the interval (12,4)(\frac{1}{2}, 4). So, the solution would be 12<x<4\frac{1}{2} < x < 4.

  2. If the inequality were 2x2βˆ’9x+4β‰₯02x^2 - 9x + 4 \geq 0, we would include the roots in the solution. The solution would be x≀12x \leq \frac{1}{2} or xβ‰₯4x \geq 4. In interval notation, this is (βˆ’βˆž,12]βˆͺ[4,∞)(-\infty, \frac{1}{2}] \cup [4, \infty).

  3. If the inequality were 2x2βˆ’9x+4≀02x^2 - 9x + 4 \leq 0, we would include the roots and look for the negative interval. The solution would be 12≀x≀4\frac{1}{2} \leq x \leq 4.

Conclusion

Solving quadratic inequalities might seem daunting at first, but by breaking it down into these four steps, it becomes much more manageable. Remember, the key is to:

  1. Rewrite the inequality in standard form.
  2. Find the roots of the corresponding quadratic equation.
  3. Create a sign chart.
  4. Determine the solution based on the sign chart and the original inequality.

With practice, you'll become a pro at solving quadratic inequalities. And remember, these skills are not just useful in math class; they have applications in various real-world scenarios.

For further learning and practice, you can check out resources like Khan Academy's Quadratic Inequalities section, which provides excellent explanations and practice problems.