Solving Absolute Value And Quadratic Equations And Inequalities
This article provides a comprehensive guide to solving absolute value equations and inequalities. We will explore various techniques and methods to tackle different types of problems, including those involving linear, quadratic, and rational expressions within the absolute value. Through detailed explanations and step-by-step solutions, you'll gain a strong understanding of how to solve these equations and inequalities effectively. Understanding absolute value equations and inequalities is crucial in various fields of mathematics and its applications. This guide aims to equip you with the necessary skills and knowledge to confidently solve a wide range of problems.
1. Solving the Absolute Value Equation |(4λ - 2)/3| = 0
To solve the absolute value equation |(4λ - 2)/3| = 0, we need to understand that the absolute value of an expression is zero if and only if the expression itself is zero. In this case, the expression inside the absolute value is (4λ - 2)/3. Therefore, we set this expression equal to zero and solve for λ.
The equation becomes:
(4λ - 2)/3 = 0
To solve this, we first multiply both sides of the equation by 3 to eliminate the fraction:
3 * (4λ - 2)/3 = 3 * 0
This simplifies to:
4λ - 2 = 0
Next, we add 2 to both sides of the equation to isolate the term with λ:
4λ - 2 + 2 = 0 + 2
This gives us:
4λ = 2
Finally, we divide both sides by 4 to solve for λ:
4λ / 4 = 2 / 4
This simplifies to:
λ = 1/2
Thus, the solution to the absolute value equation |(4λ - 2)/3| = 0 is λ = 1/2. We can verify this solution by substituting λ = 1/2 back into the original equation:
|(4*(1/2) - 2)/3| = |(2 - 2)/3| = |0/3| = |0| = 0
Since the equation holds true, our solution is correct. Understanding this method of solving absolute value equations is fundamental. The key principle is recognizing that the absolute value of an expression is zero only when the expression inside the absolute value is zero. This approach simplifies the problem to solving a basic algebraic equation. In this specific case, we dealt with a linear expression inside the absolute value, making the solution straightforward. However, this same principle applies to more complex expressions as well. The steps involved – isolating the absolute value, setting the expression inside to zero, and solving for the variable – form the basis for tackling various absolute value equations. Moreover, verifying the solution by substituting it back into the original equation is a crucial step to ensure accuracy. This process not only confirms the solution but also reinforces the understanding of the problem-solving technique. In essence, this question serves as a foundational example for understanding how to approach absolute value equations, highlighting the importance of breaking down the problem into manageable steps and applying basic algebraic principles.
2. Solving the Absolute Value Equation |25/(3 - λ)| = 0
When tackling the absolute value equation |25/(3 - λ)| = 0, we need to apply the same principle as in the previous example: the absolute value of an expression is zero if and only if the expression itself is zero. In this case, the expression inside the absolute value is 25/(3 - λ). Thus, we set this expression equal to zero and solve for λ.
The equation is:
|25/(3 - λ)| = 0
This implies that:
25/(3 - λ) = 0
Now, we need to solve this rational equation. A fraction is equal to zero if and only if its numerator is zero. The numerator here is 25, which is a constant and not equal to zero. This means that there is no value of λ that can make the fraction equal to zero. Therefore, there is no solution to this equation.
It's important to note that the denominator (3 - λ) cannot be zero, as division by zero is undefined. This would occur if λ = 3. However, since the numerator is a non-zero constant, the fraction can never be zero, regardless of the value of λ (as long as λ ≠ 3).
Therefore, the absolute value equation |25/(3 - λ)| = 0 has no solution. This problem highlights an important consideration when dealing with rational expressions within absolute values: the numerator must be zero for the entire expression to be zero. In this instance, the numerator being a non-zero constant immediately indicates that there is no solution. It is also crucial to check for values that make the denominator zero, as these values would make the expression undefined. In this case, λ = 3 would make the denominator zero, further reinforcing the conclusion that there is no solution. This type of problem emphasizes the importance of understanding the properties of fractions and how they interact with absolute values. Recognizing that a fraction can only be zero if its numerator is zero is a key concept. This understanding simplifies the process of solving absolute value equations involving rational expressions, as it allows us to quickly identify cases where no solution exists. Moreover, it reinforces the need to consider the domain of the variable, ensuring that the denominator is not zero, which is a fundamental aspect of working with rational expressions. By analyzing the structure of the equation, we can efficiently determine whether a solution is possible, saving time and effort in the problem-solving process. This approach is invaluable in more complex scenarios where the expression inside the absolute value is more intricate, and a systematic method is necessary to arrive at the correct conclusion.
3. Solving the Absolute Value Equation |2 - 3/(2 - x)| = 0
To solve the absolute value equation |2 - 3/(2 - x)| = 0, we once again rely on the principle that the absolute value of an expression is zero if and only if the expression itself is zero. In this case, the expression inside the absolute value is 2 - 3/(2 - x). Thus, we set this expression equal to zero and solve for x.
The equation becomes:
|2 - 3/(2 - x)| = 0
This implies that:
2 - 3/(2 - x) = 0
To solve this equation, we first need to eliminate the fraction. We can do this by adding 3/(2 - x) to both sides of the equation:
2 = 3/(2 - x)
Next, we multiply both sides by (2 - x) to get rid of the denominator:
2 * (2 - x) = 3
Expanding the left side, we get:
4 - 2x = 3
Now, we subtract 4 from both sides to isolate the term with x:
-2x = 3 - 4
-2x = -1
Finally, we divide both sides by -2 to solve for x:
x = (-1) / (-2)
x = 1/2
However, we need to check if this solution is valid by ensuring that the denominator (2 - x) in the original expression is not zero. If x = 1/2, then (2 - x) = 2 - 1/2 = 3/2, which is not zero. Therefore, the solution x = 1/2 is valid.
To verify the solution, we substitute x = 1/2 back into the original equation:
|2 - 3/(2 - 1/2)| = |2 - 3/(3/2)| = |2 - 3 * (2/3)| = |2 - 2| = |0| = 0
Since the equation holds true, our solution is correct. This problem introduces the additional step of dealing with a rational expression within the absolute value, which requires eliminating the fraction before solving for the variable. The process involves algebraic manipulation to isolate x, including multiplying by the denominator and simplifying the resulting equation. Crucially, it also highlights the importance of checking for extraneous solutions. Because we are dealing with a rational expression, we need to ensure that our solution does not make the denominator zero, which would render the expression undefined. This check is a critical step in solving rational equations. The verification step, where we substitute the solution back into the original equation, provides an additional layer of assurance that our solution is correct. This methodical approach, which includes solving the equation, checking for extraneous solutions, and verifying the answer, is a cornerstone of solving more complex mathematical problems. In this case, the combination of absolute value and rational expression requires careful attention to detail and a thorough understanding of algebraic principles. By systematically applying these principles, we can successfully solve the equation and ensure the validity of our solution.
4. Solving the Absolute Value Equation |2 - 3/(2 - x)| = 4
To solve the absolute value equation |2 - 3/(2 - x)| = 4, we need to consider two cases, since the expression inside the absolute value can be either 4 or -4. This is because the absolute value of both 4 and -4 is 4.
Case 1: 2 - 3/(2 - x) = 4
First, we subtract 2 from both sides of the equation:
-3/(2 - x) = 2
Next, we multiply both sides by (2 - x) to eliminate the denominator:
-3 = 2(2 - x)
Expanding the right side, we get:
-3 = 4 - 2x
Now, we add 2x to both sides and add 3 to both sides to isolate the term with x:
2x = 7
Finally, we divide both sides by 2 to solve for x:
x = 7/2
Case 2: 2 - 3/(2 - x) = -4
First, we subtract 2 from both sides of the equation:
-3/(2 - x) = -6
Next, we multiply both sides by (2 - x) to eliminate the denominator:
-3 = -6(2 - x)
Expanding the right side, we get:
-3 = -12 + 6x
Now, we subtract -12 to both sides and divide by 6:
9 = 6x
x = 9/6 = 3/2
Now, we need to check if these solutions are valid by ensuring that the denominator (2 - x) in the original expression is not zero.
For x = 7/2, (2 - x) = 2 - 7/2 = -3/2, which is not zero.
For x = 3/2, (2 - x) = 2 - 3/2 = 1/2, which is also not zero.
Therefore, both solutions are valid.
To verify the solutions, we substitute x = 7/2 and x = 3/2 back into the original equation:
For x = 7/2:
|2 - 3/(2 - 7/2)| = |2 - 3/(-3/2)| = |2 - 3*(-2/3)| = |2 + 2| = |4| = 4
For x = 3/2:
|2 - 3/(2 - 3/2)| = |2 - 3/(1/2)| = |2 - 3*2| = |2 - 6| = |-4| = 4
Since the equation holds true for both solutions, our solutions are correct. This problem introduces the crucial concept of solving absolute value equations by considering two separate cases. The reason for this is the fundamental property of absolute values: |a| = b implies that a = b or a = -b. This means that the expression inside the absolute value can be either the positive or the negative value of the number on the other side of the equation. In this specific case, |2 - 3/(2 - x)| = 4 means that either 2 - 3/(2 - x) = 4 or 2 - 3/(2 - x) = -4. Solving each of these equations separately leads to two potential solutions for x. The process of solving each case involves algebraic manipulation similar to the previous example, including eliminating the fraction and isolating x. However, the added complexity of dealing with two separate equations doubles the work required to arrive at the solutions. Furthermore, checking for extraneous solutions remains a critical step. As before, we need to ensure that our solutions do not make the denominator in the original expression zero. This check ensures the validity of our solutions. The verification step, where we substitute each solution back into the original absolute value equation, is particularly important in this type of problem. It confirms that both solutions satisfy the original equation and that we have not made any errors in our algebraic manipulations. This comprehensive approach to solving absolute value equations, which includes considering both cases, solving each equation, checking for extraneous solutions, and verifying the answers, is essential for tackling more complex problems involving absolute values. It highlights the importance of understanding the underlying principles of absolute values and applying them systematically to arrive at the correct solutions.
5. Solving the Absolute Value Inequality |x - 2| < 2
To solve the absolute value inequality |x - 2| < 2, we need to understand what this inequality means in terms of distance. The absolute value |x - 2| represents the distance between x and 2 on the number line. The inequality |x - 2| < 2 means that this distance must be less than 2. This can be interpreted as x being within 2 units of 2.
This leads to two inequalities that we need to solve:
-2 < x - 2 < 2
To solve this compound inequality, we add 2 to all parts of the inequality:
-2 + 2 < x - 2 + 2 < 2 + 2
This simplifies to:
0 < x < 4
Therefore, the solution to the absolute value inequality |x - 2| < 2 is 0 < x < 4. This means that x must be greater than 0 and less than 4. We can represent this solution graphically on a number line as an open interval between 0 and 4.
To understand this solution, consider the endpoints. If x = 0, then |0 - 2| = |-2| = 2, which is not less than 2. Similarly, if x = 4, then |4 - 2| = |2| = 2, which is also not less than 2. However, any value between 0 and 4 will satisfy the inequality. For example, if x = 2, then |2 - 2| = |0| = 0, which is less than 2.
This problem illustrates the fundamental approach to solving absolute value inequalities involving the "less than" symbol. The key is to recognize that the absolute value represents distance and to translate the inequality into a compound inequality that expresses the range of values that satisfy the condition. The compound inequality, in this case, is of the form -a < expression < a, where a is the constant on the right-hand side of the original inequality. Solving this compound inequality involves performing the same operation on all parts of the inequality, in this case, adding 2. The resulting solution is an interval that represents all the values of x that are within a certain distance of a given point. In this case, the solution 0 < x < 4 represents all the values of x that are within 2 units of 2. The graphical representation of the solution on a number line provides a visual confirmation of the range of values that satisfy the inequality. The endpoints of the interval are not included in the solution because the original inequality is a strict inequality (less than, not less than or equal to). This careful analysis of the problem, which includes understanding the meaning of absolute value, translating the inequality into a compound inequality, solving the compound inequality, and interpreting the solution, is essential for tackling more complex absolute value inequalities. It highlights the importance of understanding the underlying principles of absolute values and inequalities and applying them systematically to arrive at the correct solution.
6. Solving the Absolute Value Inequality |x + 2| > 1
To solve the absolute value inequality |x + 2| > 1, we need to interpret this inequality in terms of distance, similar to the previous example. The absolute value |x + 2| represents the distance between x and -2 on the number line. The inequality |x + 2| > 1 means that this distance must be greater than 1. This can be interpreted as x being more than 1 unit away from -2.
This leads to two separate inequalities that we need to solve:
Case 1: x + 2 > 1
Subtracting 2 from both sides, we get:
x > -1
Case 2: x + 2 < -1
Subtracting 2 from both sides, we get:
x < -3
Therefore, the solution to the absolute value inequality |x + 2| > 1 is x < -3 or x > -1. This means that x must be less than -3 or greater than -1. We can represent this solution graphically on a number line as two open intervals, one extending to the left from -3 and the other extending to the right from -1.
To understand this solution, consider the boundary points. If x = -3, then |-3 + 2| = |-1| = 1, which is not greater than 1. Similarly, if x = -1, then |-1 + 2| = |1| = 1, which is also not greater than 1. However, any value less than -3 or greater than -1 will satisfy the inequality. For example, if x = -4, then |-4 + 2| = |-2| = 2, which is greater than 1. If x = 0, then |0 + 2| = |2| = 2, which is also greater than 1.
This problem illustrates the approach to solving absolute value inequalities involving the "greater than" symbol. The key is to recognize that the absolute value represents distance and to translate the inequality into two separate inequalities. The two inequalities arise because the expression inside the absolute value can be either greater than the positive value or less than the negative value of the constant on the right-hand side of the original inequality. In this case, |x + 2| > 1 leads to two inequalities: x + 2 > 1 and x + 2 < -1. Solving each of these inequalities separately leads to two separate intervals that represent the solution. The solution is the union of these two intervals. The graphical representation of the solution on a number line provides a visual confirmation of the ranges of values that satisfy the inequality. The boundary points are not included in the solution because the original inequality is a strict inequality (greater than, not greater than or equal to). This careful analysis of the problem, which includes understanding the meaning of absolute value, translating the inequality into two separate inequalities, solving each inequality, and interpreting the solution, is essential for tackling more complex absolute value inequalities. It highlights the importance of recognizing the different approaches required for "less than" and "greater than" inequalities and applying them systematically to arrive at the correct solution. Understanding the concept of distance and how it relates to absolute value is crucial for visualizing and solving these types of problems.
7. Solving the Absolute Value Inequality |x² - 2| ≤ 1
To solve the absolute value inequality |x² - 2| ≤ 1, we need to consider the two cases that arise from the absolute value. The inequality |x² - 2| ≤ 1 means that the expression x² - 2 is within 1 unit of 0. This can be written as a compound inequality:
-1 ≤ x² - 2 ≤ 1
To solve this compound inequality, we first add 2 to all parts of the inequality:
-1 + 2 ≤ x² - 2 + 2 ≤ 1 + 2
This simplifies to:
1 ≤ x² ≤ 3
Now, we have two inequalities to solve separately:
- x² ≥ 1
- x² ≤ 3
Solving x² ≥ 1:
This inequality is satisfied when x ≤ -1 or x ≥ 1. We can see this by considering the graph of y = x², which is a parabola opening upwards. The values of x for which x² is greater than or equal to 1 are those outside the interval (-1, 1).
Solving x² ≤ 3:
This inequality is satisfied when -√3 ≤ x ≤ √3. This is because the square root function is increasing, so x² is less than or equal to 3 when x is between the negative and positive square roots of 3.
Now, we need to find the intersection of the solutions to both inequalities. The solution to x² ≥ 1 is x ≤ -1 or x ≥ 1, and the solution to x² ≤ 3 is -√3 ≤ x ≤ √3. The intersection of these two solutions is the set of x values that satisfy both inequalities. This gives us two intervals:
-√3 ≤ x ≤ -1
1 ≤ x ≤ √3
Therefore, the solution to the absolute value inequality |x² - 2| ≤ 1 is -√3 ≤ x ≤ -1 or 1 ≤ x ≤ √3. This means that x must be between -√3 and -1, inclusive, or between 1 and √3, inclusive. We can represent this solution graphically on a number line as two closed intervals.
This problem introduces the complexity of dealing with a quadratic expression inside the absolute value. The initial step of translating the absolute value inequality into a compound inequality is similar to the linear case. However, the resulting inequalities involve x², which requires a different approach to solve. The inequality x² ≥ 1 is solved by considering the intervals where the square of x is greater than or equal to 1, which leads to two separate intervals. The inequality x² ≤ 3 is solved by taking the square root of both sides, which gives a single interval. The final solution is the intersection of these intervals, which represents the values of x that satisfy both inequalities. The graphical representation of the solution on a number line provides a visual confirmation of the intervals that satisfy the inequality. The endpoints of the intervals are included in the solution because the original inequality is a non-strict inequality (less than or equal to). This problem highlights the importance of understanding how to solve quadratic inequalities and how to combine the solutions from multiple inequalities to arrive at the final answer. The use of the graph of y = x² to visualize the solution to x² ≥ 1 is a helpful technique for understanding the behavior of quadratic functions and their inequalities. The methodical approach of breaking down the problem into smaller parts, solving each part separately, and then combining the solutions is essential for tackling more complex mathematical problems. This careful analysis ensures that all possible solutions are considered and that the final answer is accurate.
8. Solving the Absolute Value Inequality |12 - t| ≥ 1
To solve the absolute value inequality |12 - t| ≥ 1, we need to interpret this inequality in terms of distance, similar to the previous examples. The absolute value |12 - t| represents the distance between 12 and t on the number line. The inequality |12 - t| ≥ 1 means that this distance must be greater than or equal to 1. This can be interpreted as t being at least 1 unit away from 12.
This leads to two separate inequalities that we need to solve:
Case 1: 12 - t ≥ 1
Subtracting 12 from both sides, we get:
-t ≥ -11
Multiplying both sides by -1 (and reversing the inequality sign), we get:
t ≤ 11
Case 2: 12 - t ≤ -1
Subtracting 12 from both sides, we get:
-t ≤ -13
Multiplying both sides by -1 (and reversing the inequality sign), we get:
t ≥ 13
Therefore, the solution to the absolute value inequality |12 - t| ≥ 1 is t ≤ 11 or t ≥ 13. This means that t must be less than or equal to 11 or greater than or equal to 13. We can represent this solution graphically on a number line as two closed intervals, one extending to the left from 11 and the other extending to the right from 13.
To understand this solution, consider the boundary points. If t = 11, then |12 - 11| = |1| = 1, which is equal to 1. If t = 13, then |12 - 13| = |-1| = 1, which is also equal to 1. Any value less than 11 or greater than 13 will satisfy the inequality. For example, if t = 10, then |12 - 10| = |2| = 2, which is greater than 1. If t = 14, then |12 - 14| = |-2| = 2, which is also greater than 1.
This problem reinforces the approach to solving absolute value inequalities involving the "greater than or equal to" symbol. The key is to recognize that the absolute value represents distance and to translate the inequality into two separate inequalities. The two inequalities arise because the expression inside the absolute value can be either greater than or equal to the positive value or less than or equal to the negative value of the constant on the right-hand side of the original inequality. In this case, |12 - t| ≥ 1 leads to two inequalities: 12 - t ≥ 1 and 12 - t ≤ -1. Solving each of these inequalities separately leads to two separate intervals that represent the solution. The solution is the union of these two intervals. The graphical representation of the solution on a number line provides a visual confirmation of the ranges of values that satisfy the inequality. The boundary points are included in the solution because the original inequality is a non-strict inequality (greater than or equal to). This careful analysis of the problem, which includes understanding the meaning of absolute value, translating the inequality into two separate inequalities, solving each inequality, and interpreting the solution, is essential for tackling more complex absolute value inequalities. It highlights the importance of recognizing the different approaches required for "less than" and "greater than" inequalities and applying them systematically to arrive at the correct solution. The additional step of multiplying by -1 and reversing the inequality sign is a common source of error, so it is important to pay close attention to this step. Understanding the concept of distance and how it relates to absolute value is crucial for visualizing and solving these types of problems.
9. Solving the Inequality λ² - 4 ≤ 0
To solve the inequality λ² - 4 ≤ 0, we first recognize that this is a quadratic inequality. To solve it, we first find the roots of the corresponding quadratic equation λ² - 4 = 0. This equation can be factored as a difference of squares:
(λ - 2)(λ + 2) = 0
The roots of this equation are λ = 2 and λ = -2. These roots divide the number line into three intervals: λ < -2, -2 < λ < 2, and λ > 2. We need to test a value from each interval to determine where the inequality λ² - 4 ≤ 0 is satisfied.
Interval 1: λ < -2
Let's test λ = -3:
(-3)² - 4 = 9 - 4 = 5
Since 5 > 0, the inequality is not satisfied in this interval.
Interval 2: -2 < λ < 2
Let's test λ = 0:
(0)² - 4 = -4
Since -4 ≤ 0, the inequality is satisfied in this interval.
Interval 3: λ > 2
Let's test λ = 3:
(3)² - 4 = 9 - 4 = 5
Since 5 > 0, the inequality is not satisfied in this interval.
Since the inequality is non-strict (≤), we also include the roots λ = -2 and λ = 2 in the solution. Therefore, the solution to the inequality λ² - 4 ≤ 0 is -2 ≤ λ ≤ 2. This means that λ must be between -2 and 2, inclusive. We can represent this solution graphically on a number line as a closed interval between -2 and 2.
This problem illustrates the standard approach to solving quadratic inequalities. The first step is to find the roots of the corresponding quadratic equation. These roots are the points where the quadratic expression changes sign. The roots divide the number line into intervals, and we test a value from each interval to determine where the inequality is satisfied. The sign of the quadratic expression in each interval is determined by the signs of the factors (λ - 2) and (λ + 2). The quadratic expression is negative between the roots and positive outside the roots. The inclusion of the roots in the solution depends on whether the inequality is strict (< or >) or non-strict (≤ or ≥). In this case, the inequality is non-strict, so we include the roots in the solution. The graphical representation of the solution on a number line provides a visual confirmation of the interval that satisfies the inequality. This methodical approach, which includes finding the roots, testing intervals, and considering the type of inequality, is essential for solving quadratic inequalities. It highlights the importance of understanding the relationship between the roots of a quadratic equation and the sign of the quadratic expression. The use of a sign chart can be a helpful tool for organizing the information and determining the intervals where the inequality is satisfied. This careful analysis ensures that all possible solutions are considered and that the final answer is accurate.
10. Solving the Inequality λ² + λ - 6 > 0
To solve the inequality λ² + λ - 6 > 0, we again recognize that this is a quadratic inequality. We follow the same procedure as in the previous example: first, we find the roots of the corresponding quadratic equation λ² + λ - 6 = 0. This equation can be factored as:
(λ + 3)(λ - 2) = 0
The roots of this equation are λ = -3 and λ = 2. These roots divide the number line into three intervals: λ < -3, -3 < λ < 2, and λ > 2. We need to test a value from each interval to determine where the inequality λ² + λ - 6 > 0 is satisfied.
Interval 1: λ < -3
Let's test λ = -4:
(-4)² + (-4) - 6 = 16 - 4 - 6 = 6
Since 6 > 0, the inequality is satisfied in this interval.
Interval 2: -3 < λ < 2
Let's test λ = 0:
(0)² + (0) - 6 = -6
Since -6 < 0, the inequality is not satisfied in this interval.
Interval 3: λ > 2
Let's test λ = 3:
(3)² + (3) - 6 = 9 + 3 - 6 = 6
Since 6 > 0, the inequality is satisfied in this interval.
Since the inequality is strict (>), we do not include the roots λ = -3 and λ = 2 in the solution. Therefore, the solution to the inequality λ² + λ - 6 > 0 is λ < -3 or λ > 2. This means that λ must be less than -3 or greater than 2. We can represent this solution graphically on a number line as two open intervals, one extending to the left from -3 and the other extending to the right from 2.
This problem further illustrates the standard approach to solving quadratic inequalities. The key steps are to find the roots of the corresponding quadratic equation, divide the number line into intervals based on the roots, test a value from each interval to determine where the inequality is satisfied, and consider whether the roots are included in the solution based on the type of inequality (strict or non-strict). The factorization of the quadratic equation is a crucial step, as it allows us to easily identify the roots. The testing of values in each interval is a systematic way to determine the sign of the quadratic expression in each interval. The graphical representation of the solution on a number line provides a visual confirmation of the intervals that satisfy the inequality. The exclusion of the roots in this case is due to the strict inequality (>), which means that the quadratic expression must be strictly greater than 0, not equal to 0. This careful analysis ensures that all possible solutions are considered and that the final answer is accurate. The use of a sign chart can be a helpful tool for organizing the information and determining the intervals where the inequality is satisfied, especially for more complex quadratic inequalities. This problem reinforces the understanding of the relationship between the roots of a quadratic equation and the sign of the quadratic expression and how to use this relationship to solve inequalities.
11. Solving the Inequality (x - 1)(x + 3) ≥ 5
To solve the inequality (x - 1)(x + 3) ≥ 5, we first need to rewrite the inequality in the standard form of a quadratic inequality, which means having 0 on one side. To do this, we first expand the left side and then subtract 5 from both sides:
(x - 1)(x + 3) = x² + 3x - x - 3 = x² + 2x - 3
Now, subtract 5 from both sides:
x² + 2x - 3 - 5 ≥ 0
This simplifies to:
x² + 2x - 8 ≥ 0
Now we have a quadratic inequality in the standard form. We proceed as in the previous examples: first, we find the roots of the corresponding quadratic equation x² + 2x - 8 = 0. This equation can be factored as:
(x + 4)(x - 2) = 0
The roots of this equation are x = -4 and x = 2. These roots divide the number line into three intervals: x < -4, -4 < x < 2, and x > 2. We need to test a value from each interval to determine where the inequality x² + 2x - 8 ≥ 0 is satisfied.
Interval 1: x < -4
Let's test x = -5:
(-5)² + 2(-5) - 8 = 25 - 10 - 8 = 7
Since 7 ≥ 0, the inequality is satisfied in this interval.
Interval 2: -4 < x < 2
Let's test x = 0:
(0)² + 2(0) - 8 = -8
Since -8 < 0, the inequality is not satisfied in this interval.
Interval 3: x > 2
Let's test x = 3:
(3)² + 2(3) - 8 = 9 + 6 - 8 = 7
Since 7 ≥ 0, the inequality is satisfied in this interval.
Since the inequality is non-strict (≥), we also include the roots x = -4 and x = 2 in the solution. Therefore, the solution to the inequality (x - 1)(x + 3) ≥ 5 is x ≤ -4 or x ≥ 2. This means that x must be less than or equal to -4 or greater than or equal to 2. We can represent this solution graphically on a number line as two closed intervals, one extending to the left from -4 and the other extending to the right from 2.
This problem highlights the importance of rewriting an inequality in standard form before solving it. The initial inequality (x - 1)(x + 3) ≥ 5 needs to be transformed into the standard form x² + 2x - 8 ≥ 0 before we can apply the standard procedure for solving quadratic inequalities. This transformation involves expanding the product on the left side and then subtracting 5 from both sides. Once the inequality is in standard form, we can proceed as in the previous examples, finding the roots of the corresponding quadratic equation, dividing the number line into intervals, testing values in each interval, and considering whether the roots are included in the solution. The factorization of the quadratic equation is again a crucial step, and the testing of values in each interval is a systematic way to determine the sign of the quadratic expression. The graphical representation of the solution on a number line provides a visual confirmation of the intervals that satisfy the inequality. The inclusion of the roots in this case is due to the non-strict inequality (≥), which means that the quadratic expression can be equal to 0. This careful analysis ensures that all possible solutions are considered and that the final answer is accurate. This problem reinforces the understanding of the steps involved in solving quadratic inequalities and the importance of rewriting the inequality in standard form before applying the solution procedure.
12. Solving the Inequality x² + 4x + 5 > 0
To solve the inequality x² + 4x + 5 > 0, we follow the same general approach as with other quadratic inequalities. First, we consider the corresponding quadratic equation x² + 4x + 5 = 0. To find the roots of this equation, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = 5. Plugging these values into the quadratic formula, we get:
x = (-4 ± √(4² - 4 * 1 * 5)) / (2 * 1)
x = (-4 ± √(16 - 20)) / 2
x = (-4 ± √(-4)) / 2
Since the discriminant (the value inside the square root) is negative, the quadratic equation has no real roots. This means that the parabola represented by the quadratic expression x² + 4x + 5 does not intersect the x-axis. Since the coefficient of x² is positive (a = 1), the parabola opens upwards. This means that the quadratic expression is either always positive or always negative.
To determine whether the expression is always positive or always negative, we can test any value of x. Let's test x = 0:
(0)² + 4(0) + 5 = 5
Since 5 > 0, the quadratic expression is positive when x = 0. Since the parabola does not intersect the x-axis and opens upwards, the expression is positive for all real values of x.
Therefore, the solution to the inequality x² + 4x + 5 > 0 is all real numbers. This means that any value of x will satisfy the inequality. We can represent this solution graphically on a number line as the entire number line.
This problem introduces a scenario where the quadratic equation has no real roots. This occurs when the discriminant of the quadratic formula is negative. In this case, the parabola represented by the quadratic expression does not intersect the x-axis, which means that the expression is either always positive or always negative. To determine which is the case, we can test any value of x. If the expression is positive for that value of x, then it is positive for all values of x. If it is negative, then it is negative for all values of x. In this case, the expression is positive for x = 0, so it is positive for all real values of x. The graphical representation of the solution on a number line is the entire number line, which indicates that all real numbers satisfy the inequality. This problem highlights the importance of considering the discriminant when solving quadratic inequalities. If the discriminant is negative, then the quadratic equation has no real roots, and the solution to the inequality is either all real numbers or no real numbers, depending on the sign of the quadratic expression. This careful analysis ensures that we arrive at the correct solution, even in cases where the quadratic equation has no real roots. Understanding the relationship between the discriminant and the nature of the roots is crucial for solving quadratic inequalities effectively.
13. Solving the Inequality x² + 5 < -3x
To solve the inequality x² + 5 < -3x, we first need to rewrite the inequality in the standard form of a quadratic inequality, which means having 0 on one side. To do this, we add 3x to both sides:
x² + 3x + 5 < 0
Now we have a quadratic inequality in the standard form. We proceed as in the previous examples: first, we find the roots of the corresponding quadratic equation x² + 3x + 5 = 0. To find the roots, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 1, b = 3, and c = 5. Plugging these values into the quadratic formula, we get:
x = (-3 ± √(3² - 4 * 1 * 5)) / (2 * 1)
x = (-3 ± √(9 - 20)) / 2
x = (-3 ± √(-11)) / 2
Since the discriminant (the value inside the square root) is negative, the quadratic equation has no real roots. This means that the parabola represented by the quadratic expression x² + 3x + 5 does not intersect the x-axis. Since the coefficient of x² is positive (a = 1), the parabola opens upwards. This means that the quadratic expression is either always positive or always negative.
To determine whether the expression is always positive or always negative, we can test any value of x. Let's test x = 0:
(0)² + 3(0) + 5 = 5
Since 5 > 0, the quadratic expression is positive when x = 0. Since the parabola does not intersect the x-axis and opens upwards, the expression is positive for all real values of x. Therefore, the inequality x² + 3x + 5 < 0 has no real solutions, because the quadratic expression is always positive and never less than 0.
This problem further illustrates the scenario where the quadratic equation has no real roots. The steps involved in solving the inequality are the same as in the previous example: rewrite the inequality in standard form, find the roots of the corresponding quadratic equation using the quadratic formula, and determine the sign of the quadratic expression. The negative discriminant indicates that there are no real roots, and the positive coefficient of x² indicates that the parabola opens upwards. The testing of a value of x (in this case, x = 0) confirms that the quadratic expression is always positive. Therefore, the inequality x² + 3x + 5 < 0 has no solution, as the expression is never less than 0. This careful analysis ensures that we arrive at the correct solution, even in cases where there are no real solutions. The graphical representation of the quadratic expression as a parabola that does not intersect the x-axis provides a visual confirmation of the absence of solutions. This problem reinforces the understanding of the relationship between the discriminant, the roots of a quadratic equation, and the sign of the quadratic expression, and how to use this knowledge to solve inequalities. Understanding these concepts is crucial for tackling a wide range of quadratic inequalities and ensuring accurate solutions.
In conclusion, this comprehensive guide has provided a thorough exploration of solving various types of absolute value equations and inequalities, as well as quadratic inequalities. From basic absolute value equations to more complex scenarios involving rational and quadratic expressions, we have covered the fundamental principles and techniques required to tackle these problems effectively. The key concepts emphasized include understanding the meaning of absolute value in terms of distance, translating absolute value equations and inequalities into appropriate algebraic equations and inequalities, solving quadratic equations and inequalities, checking for extraneous solutions, and interpreting the solutions graphically. By mastering these concepts and techniques, you will be well-equipped to solve a wide range of mathematical problems involving absolute values and quadratic expressions. The step-by-step solutions and detailed explanations provided in this guide serve as a valuable resource for students and anyone seeking to deepen their understanding of these important mathematical topics. Remember to practice regularly and apply these techniques to different types of problems to solidify your knowledge and skills. With consistent effort and a solid grasp of the underlying principles, you can confidently solve even the most challenging absolute value and quadratic inequalities.
