Solving A System Of Inequalities: A Step-by-Step Guide
Hey guys! Today, we're diving into the fascinating world of algebra to tackle a system of inequalities. Specifically, we're going to solve the following system:
{ x^2 + x - 2 ≤ 0,
x^2 + 2x - 8 ≥ 0 }
Don't worry if it looks intimidating at first! We'll break it down step-by-step, making sure you understand the logic behind each move. So, grab your pencils and let's get started!
Understanding Inequalities and Systems
Before we jump into the solution, let's quickly recap what inequalities are and what it means to solve a system of them. Inequalities are mathematical statements that compare two expressions using symbols like < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to). Unlike equations, which have specific solutions, inequalities often have a range of solutions. A system of inequalities is simply a set of two or more inequalities that we need to solve simultaneously. This means we're looking for the values of x that satisfy all the inequalities in the system at the same time.
When we talk about solving inequalities, especially quadratic inequalities like these, the core idea is to find the intervals on the number line where the inequality holds true. This often involves finding the roots of the quadratic expressions and then testing values within the intervals defined by those roots. Think of it like a puzzle – we're piecing together the sections of the number line that fit our conditions. We'll use a combination of algebraic manipulation and a bit of visual thinking to nail this down. So, let’s get our hands dirty and dive into the actual solving process!
Step 1: Solve the First Inequality (x² + x - 2 ≤ 0)
Okay, let's tackle the first inequality: x² + x - 2 ≤ 0. Our goal here is to find the values of x that make this statement true. The best way to do this is by factoring the quadratic expression.
Factoring the Quadratic
We need to find two numbers that multiply to -2 and add up to 1 (the coefficient of the x term). Those numbers are 2 and -1. So, we can factor the expression as follows:
x² + x - 2 = (x + 2)(x - 1)
Now our inequality looks like this:
(x + 2)(x - 1) ≤ 0
Finding the Critical Points
The critical points are the values of x that make the expression equal to zero. These points are important because they divide the number line into intervals where the expression is either positive or negative. To find them, we set each factor equal to zero:
x + 2 = 0 => x = -2
x - 1 = 0 => x = 1
So, our critical points are x = -2 and x = 1.
Testing Intervals
These critical points divide the number line into three intervals: (-∞, -2), (-2, 1), and (1, ∞). We need to test a value from each interval in the inequality (x + 2)(x - 1) ≤ 0 to see if it holds true.
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Interval (-∞, -2): Let's test x = -3
(-3 + 2)(-3 - 1) = (-1)(-4) = 44 is not less than or equal to 0, so this interval is not part of the solution.
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Interval (-2, 1): Let's test x = 0
(0 + 2)(0 - 1) = (2)(-1) = -2-2 is less than or equal to 0, so this interval is part of the solution.
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Interval (1, ∞): Let's test x = 2
(2 + 2)(2 - 1) = (4)(1) = 44 is not less than or equal to 0, so this interval is not part of the solution.
Since the inequality is less than or equal to zero, we also include the critical points themselves in the solution. Therefore, the solution to the first inequality is -2 ≤ x ≤ 1. We can write this in interval notation as [-2, 1].
Step 2: Solve the Second Inequality (x² + 2x - 8 ≥ 0)
Now, let's move on to the second inequality: x² + 2x - 8 ≥ 0. We'll follow the same process as before: factoring the quadratic, finding critical points, and testing intervals.
Factoring the Quadratic
We need to find two numbers that multiply to -8 and add up to 2. Those numbers are 4 and -2. So, we can factor the expression as:
x² + 2x - 8 = (x + 4)(x - 2)
Our inequality now looks like this:
(x + 4)(x - 2) ≥ 0
Finding the Critical Points
Set each factor equal to zero:
x + 4 = 0 => x = -4
x - 2 = 0 => x = 2
Our critical points are x = -4 and x = 2.
Testing Intervals
These critical points divide the number line into three intervals: (-∞, -4), (-4, 2), and (2, ∞). Let's test a value from each interval in the inequality (x + 4)(x - 2) ≥ 0.
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Interval (-∞, -4): Let's test x = -5
(-5 + 4)(-5 - 2) = (-1)(-7) = 77 is greater than or equal to 0, so this interval is part of the solution.
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Interval (-4, 2): Let's test x = 0
(0 + 4)(0 - 2) = (4)(-2) = -8-8 is not greater than or equal to 0, so this interval is not part of the solution.
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Interval (2, ∞): Let's test x = 3
(3 + 4)(3 - 2) = (7)(1) = 77 is greater than or equal to 0, so this interval is part of the solution.
Since the inequality is greater than or equal to zero, we include the critical points in the solution. Therefore, the solution to the second inequality is x ≤ -4 or x ≥ 2. In interval notation, this is (-∞, -4] ∪ [2, ∞).
Step 3: Find the Intersection of the Solutions
Alright, we've solved each inequality separately. Now comes the crucial part: finding the intersection of the solutions. Remember, we need to find the values of x that satisfy both inequalities simultaneously. This means we need to find where the two solution sets overlap.
- Solution to the first inequality: [-2, 1]
- Solution to the second inequality: (-∞, -4] ∪ [2, ∞)
Let's visualize this on a number line. Imagine shading the interval [-2, 1] on one line and the intervals (-∞, -4] and [2, ∞) on another line. The intersection is the region where the shading overlaps.
When we look for the overlap, we see that there is no overlap between the two solution sets. The interval [-2, 1] lies entirely between -2 and 1, while the solution to the second inequality consists of values less than or equal to -4 and values greater than or equal to 2. There's simply no common ground.
The Final Solution
Since there's no overlap between the solution sets of the two inequalities, the system of inequalities has no solution. We can represent this using the empty set symbol, which looks like this: ∅.
So, the final answer is:
x ∈ ∅
Wrapping Up
And there you have it! We've successfully solved a system of quadratic inequalities. Remember, the key is to break down the problem into smaller steps: factor the quadratics, find the critical points, test intervals, and then find the intersection of the solutions. Don't be afraid to draw a number line to visualize the intervals – it can make the process much clearer.
Solving systems of inequalities might seem tricky at first, but with practice, you'll become a pro. Keep practicing, and you'll be conquering complex algebraic problems in no time! Keep up the great work, guys! You've got this!