Solving Quadratic Inequality: What Are The Solution Sets?

Understanding quadratic inequalities and their solution sets is a fundamental concept in mathematics. In this comprehensive guide, we'll explore how to determine the solution sets for quadratic inequalities, specifically focusing on inequalities in the form $f(x) extgreater{=} 0$. We'll break down the key concepts, provide step-by-step explanations, and illustrate with examples to ensure a clear understanding. Whether you're a student tackling algebra or just looking to brush up on your math skills, this article will provide you with the knowledge you need.

Understanding Quadratic Inequalities

Before diving into the solution sets, let's first define what a quadratic inequality is. A quadratic inequality is a mathematical expression that compares a quadratic function to another value, which is often zero. The general form of a quadratic inequality is $ax^2 + bx + c extgreater 0$, $ax^2 + bx + c extless 0$, $ax^2 + bx + c extgreater{=} 0$, or $ax^2 + bx + c extless{=} 0$, where a, b, and c are constants, and a is not equal to zero. The solutions to these inequalities are the values of x that make the inequality true. To effectively solve these inequalities, it's essential to understand the behavior of quadratic functions and their graphical representations.

Now, let's talk about how to tackle these inequalities. When you're dealing with $f(x) extgreater{=} 0$, you're essentially trying to find all the x values that make the quadratic function greater than or equal to zero. This means you're looking for the regions on the graph of the quadratic function that lie above the x-axis (where $f(x)$ is positive) and the points where the graph touches the x-axis (where $f(x)$ is zero). This involves a combination of algebraic techniques and graphical analysis. Understanding the roots of the quadratic equation and the shape of the parabola is crucial in determining the solution set. So, let's delve deeper into the methods for finding these solutions.

The solution set represents all the x values that satisfy the inequality. It can be expressed in several ways, including interval notation, set notation, or graphically on a number line. Each method offers a unique way to visualize and represent the solutions. For example, if the solution set is all real numbers greater than or equal to a certain value, we can represent it using interval notation as $[a, ext{\infty})$, where a is the lower bound. Set notation would represent this as ${x \mid x \geq a}$, and graphically, it would be a shaded region on the number line starting from a and extending to the right. Understanding these different representations helps in interpreting and communicating the solutions effectively.

Key Steps to Solve Quadratic Inequalities

To solve quadratic inequalities, follow these key steps:

1. Rewrite the Inequality: Start by rearranging the inequality so that one side is zero. This puts the inequality in a standard form, making it easier to analyze. For example, if you have $ax^2 + bx + c extgreater d$, rewrite it as $ax^2 + bx + (c - d) extgreater 0$. This step simplifies the process of finding critical points and testing intervals.

2. Find the Roots: Determine the roots (or zeros) of the corresponding quadratic equation $f(x) = 0$. These roots are the points where the parabola intersects the x-axis. You can find the roots by factoring, using the quadratic formula, or completing the square. The roots are crucial because they divide the number line into intervals that we'll test in the next step.

3. Identify the Intervals: The roots divide the number line into intervals. These intervals represent different ranges of x values where the quadratic function will either be positive or negative. For example, if the roots are -2 and 3, the intervals will be $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. Understanding these intervals is essential for determining where the inequality holds true.

4. Test Each Interval: Choose a test value within each interval and substitute it into the original inequality. This will tell you whether the quadratic function is positive or negative in that interval. If the test value satisfies the inequality, then all values in that interval are part of the solution set. This step is the heart of solving quadratic inequalities, as it helps us identify the ranges of x values that satisfy the condition.

5. Write the Solution Set: Based on the interval tests, write the solution set using interval notation, set notation, or a graph. Be mindful of whether the inequality includes equality ($\textgreater{=}$ or $\textless{=}$), in which case the roots themselves are part of the solution set. Representing the solution set accurately is the final step in the process, ensuring that all values satisfying the inequality are included.

Possible Solution Sets for $f(x) extgreater{=} 0$

Now, let's consider the possible solution sets for the quadratic inequality $f(x) extgreater{=} 0$. The solution set depends on the discriminant of the quadratic equation and the leading coefficient.

Case 1: Two Distinct Real Roots

If the quadratic equation has two distinct real roots, say $x_1$ and $x_2$ (where $x_1 extless x_2$), and the leading coefficient a is positive, then the parabola opens upwards. In this case, $f(x) extgreater{=} 0$ for $x \textless{=} x_1$ or $x extgreater{=} x_2$. The solution set is $(-\infty, x_1] \cup [x_2, \infty)$. If a is negative, the parabola opens downwards, and $f(x) extgreater{=} 0$ for $x_1 \textless{=} x \textless{=} x_2$. The solution set is $[x_1, x_2]$. Understanding the direction in which the parabola opens is crucial for determining the solution set.

For example, consider the inequality $x^2 - 5x + 6 extgreater{=} 0$. The roots are $x_1 = 2$ and $x_2 = 3$, and the leading coefficient is positive. The solution set is $(-\infty, 2] \cup [3, \infty)$. This means that all values of x less than or equal to 2, and all values greater than or equal to 3, will satisfy the inequality. Graphically, the parabola lies above or on the x-axis in these intervals. The union of these intervals forms the complete solution set.

Case 2: One Real Root (Repeated Root)

If the quadratic equation has one real root (a repeated root), say x₀, the solution set depends on the leading coefficient a. If a is positive, the parabola touches the x-axis at x₀ and opens upwards. In this case, $f(x) extgreater{=} 0$ for all x, so the solution set is ${x \mid x \in R }$, which means all real numbers. If a is negative, the parabola touches the x-axis at x₀ and opens downwards, and $f(x) extgreater{=} 0$ only at x = x₀. The solution set is ${x \mid x = x_0}$. The nature of the roots significantly influences the shape and position of the parabola, and thus, the solution set.

For instance, let's look at the inequality $(x - 2)^2 extgreater{=} 0$. The repeated root is x = 2, and the leading coefficient is positive. The solution set is all real numbers, because the parabola opens upwards and touches the x-axis at x = 2. This means that any real number substituted into the inequality will make it true. The parabola is either above the x-axis or touches it at the vertex, ensuring that the inequality holds for all x.

Case 3: No Real Roots

If the quadratic equation has no real roots, the solution set depends on the leading coefficient a. If a is positive, the parabola opens upwards and never intersects the x-axis. In this case, $f(x) extgreater{=} 0$ for all x, so the solution set is ${x \mid x \in R }$. If a is negative, the parabola opens downwards and never intersects the x-axis. In this case, $f(x) extgreater{=} 0$ has no solution, so the solution set is $\varnothing$, the empty set. The absence of real roots implies that the parabola does not cross the x-axis, which drastically affects the solution set.

Consider the inequality $x^2 + 1 extgreater{=} 0$. There are no real roots because the discriminant is negative. The leading coefficient is positive, so the parabola opens upwards and never intersects the x-axis. The solution set is all real numbers, because $x^2 + 1$ is always positive. This is a classic example of a quadratic inequality with no real roots, leading to a solution set that encompasses all real numbers. On the other hand, if we had $-x^2 - 1 extgreater{=} 0$, the solution set would be the empty set, as the parabola opens downwards and is always below the x-axis.

Examples and Scenarios

To solidify your understanding, let's walk through some examples and scenarios. These examples will illustrate how to apply the steps outlined earlier and how to interpret the results in different contexts.

Example 1: $x^2 - 4x + 3 extgreater{=} 0$

1. Rewrite: The inequality is already in the form $f(x) extgreater{=} 0$.
2. Find the Roots: Factor the quadratic equation $x^2 - 4x + 3 = 0$ to get $(x - 1)(x - 3) = 0$. The roots are x₁ = 1 and x₂ = 3.
3. Identify the Intervals: The intervals are $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.
4. Test Each Interval:
   • For $(-\infty, 1)$, test x = 0: $(0)^2 - 4(0) + 3 = 3 extgreater{=} 0$ (True)
   • For $(1, 3)$, test x = 2: $(2)^2 - 4(2) + 3 = -1 \textless 0$ (False)
   • For $(3, \infty)$, test x = 4: $(4)^2 - 4(4) + 3 = 3 extgreater{=} 0$ (True)
5. Write the Solution Set: The solution set is $(-\infty, 1] \cup [3, \infty)$.

This example showcases how to systematically find the solution set by identifying roots, testing intervals, and combining the results. The solution set represents all the x values for which the parabola is above or on the x-axis. The inclusion of the roots (1 and 3) is due to the