Solving The Inequality (x+1)(x-2):|x|-1≥0: A Detailed Guide

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Hey guys! Today, we're diving deep into the world of inequalities to tackle a particularly interesting one: (x+1)(x-2):|x|-1≥0. This problem combines polynomial expressions with absolute values, making it a fantastic exercise for honing our algebraic skills. We'll break it down step-by-step, ensuring you understand not just the solution, but also the why behind each step. So, grab your thinking caps, and let's get started!

Understanding the Inequality

Before we jump into the solution, let's make sure we understand what we're dealing with. The inequality (x+1)(x-2):|x|-1≥0 involves a rational expression where the numerator is a product of two linear factors, (x+1) and (x-2), and the denominator involves the absolute value of x, |x|, minus 1. The goal is to find all values of x that satisfy this inequality, meaning the expression on the left-hand side is greater than or equal to zero.

The presence of the absolute value is a key aspect here. Remember, the absolute value of a number is its distance from zero, so |x| is equal to x if x is non-negative and -x if x is negative. This means we'll need to consider two cases: when x is positive or zero, and when x is negative. Additionally, we need to be mindful of the denominator, |x|-1, as it cannot be equal to zero (division by zero is a big no-no in mathematics!). This gives us our first critical point to consider: |x| cannot be 1, meaning x cannot be 1 or -1.

When you're looking at inequalities like this, it's super important to consider the critical points. These are the values of x that make either the numerator or the denominator equal to zero, or that make the denominator undefined. Identifying these points is the first big step to solving the inequality. In our case, the critical points arise from the factors in the numerator (x+1 and x-2) and the denominator (|x|-1).

Step-by-Step Solution

Let's break down the solution into manageable steps. This will help us tackle the inequality systematically and avoid any confusion.

1. Identify Critical Points

As we discussed, critical points are the values of x that make the numerator zero or the denominator zero or undefined.

  • Numerator: The numerator (x+1)(x-2) is zero when x = -1 or x = 2.
  • Denominator: The denominator |x|-1 is zero when |x| = 1, which means x = 1 or x = -1. It's crucial to note that x = -1 appears as a critical point for both the numerator and the denominator, which we'll need to consider carefully.
  • Additionally, the denominator is undefined when |x| - 1 = 0, so x cannot be 1 or -1. These values must be excluded from our final solution.

Therefore, our critical points are x = -1, x = 1, and x = 2. These points divide the number line into intervals, which we'll analyze in the next step.

2. Consider Cases Based on Absolute Value

Because of the absolute value, we need to consider two main cases:

  • Case 1: x ≥ 0

    When x is greater than or equal to 0, |x| is simply x. So, our inequality becomes: (x+1)(x-2)/(x-1) ≥ 0

  • Case 2: x < 0

    When x is less than 0, |x| is -x. Our inequality transforms into: (x+1)(x-2)/(-x-1) ≥ 0

Now, we have two simpler inequalities to solve, each within a specific range of x values. This is a common strategy when dealing with absolute values – break the problem into cases that eliminate the absolute value signs.

3. Analyze Intervals for Each Case

For each case, we'll create a sign chart to analyze the intervals created by our critical points. A sign chart helps us determine the sign of the expression in each interval.

  • Case 1: x ≥ 0, Inequality: (x+1)(x-2)/(x-1) ≥ 0

    The relevant critical points in this case are x = 1 and x = 2 (since we're only considering x ≥ 0). These points divide the non-negative number line into three intervals: [0, 1), (1, 2), and (2, ∞).

    Interval Test Value x+1 x-2 x-1 (x+1)(x-2)/(x-1) Sign Included?
    [0, 1) 0.5 + - - + + Yes
    (1, 2) 1.5 + - + - - No
    (2, ∞) 3 + + + + + Yes

    So, for Case 1, the solution intervals are [0, 1) and [2, ∞).

  • Case 2: x < 0, Inequality: (x+1)(x-2)/(-x-1) ≥ 0

    Here, the relevant critical point is x = -1. The intervals to consider are (-∞, -1) and (-1, 0).

    Interval Test Value x+1 x-2 -x-1 (x+1)(x-2)/(-x-1) Sign Included?
    (-∞, -1) -2 - - + + + Yes
    (-1, 0) -0.5 + - - + + Yes

    For Case 2, the solution intervals are (-∞, -1) and (-1, 0). However, we need to remember that x cannot be -1, as it makes the denominator zero. So, we express the solution as (-∞, -1) ∪ (-1, 0).

This interval analysis is super important. It's where we determine where our inequality actually holds true. By plugging in test values, we can see the sign of each factor and, therefore, the sign of the entire expression. This makes it clear which intervals satisfy our original condition of being greater than or equal to zero.

4. Combine Solutions

Finally, we combine the solutions from both cases. Remember to exclude x = 1 and x = -1, as they make the denominator zero.

  • From Case 1: [0, 1) ∪ [2, ∞)
  • From Case 2: (-∞, -1) ∪ (-1, 0)

Combining these, we get the overall solution:

(-∞, -1) ∪ (-1, 1) ∪ [2, ∞)

This is our final answer! It represents all the values of x that satisfy the original inequality. Remember to always double-check your solution by plugging in values from your solution set into the original inequality to ensure they hold true.

Visualizing the Solution

Sometimes, it helps to visualize the solution on a number line. We'll mark our critical points and shade the intervals that are part of the solution. We'll use open circles for points that are excluded (like -1 and 1) and closed brackets for points that are included (like 2).

<-------------------------------------------------------->
(      )   (      )          [------------------------>
(-∞---- -1 ---- 1 ----------- 2 ---- ∞)

This visual representation makes it clear which intervals are included in our solution set.

Key Takeaways

  • Critical Points are Key: Identifying critical points is the foundation of solving inequalities.
  • Consider Cases for Absolute Values: Break down the problem into cases based on the absolute value's definition.
  • Sign Charts are Your Friend: Use sign charts to analyze intervals and determine the sign of the expression.
  • Combine Solutions Carefully: Ensure you're combining solutions from all cases and excluding any values that make the denominator zero.
  • Always Double-Check: Verify your solution by plugging in values into the original inequality.

Practice Makes Perfect

Solving inequalities like this can seem daunting at first, but with practice, you'll become more comfortable with the process. Try tackling similar problems, and don't be afraid to make mistakes – they're part of the learning process!

Remember, the key is to break the problem down into smaller, manageable steps. By understanding the underlying principles and applying them systematically, you can conquer even the most challenging inequalities. Keep practicing, and you'll become a pro in no time!

So there you have it, guys! We've successfully navigated through the solution of the inequality (x+1)(x-2):|x|-1≥0. I hope this detailed guide has been helpful and has boosted your confidence in tackling similar problems. Keep up the great work, and happy solving!