Chapter 12: Problem 94
Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(x^{2}-3 x<6+2 x\)
Short Answer
Expert verified
The solution to the inequality is \( -1 < x < 6 \)
Step by step solution
01
- Move all terms to one side
First, we need to move all terms to one side of the inequality to set it to zero. Subtract both 6 and 2x from both sides of the inequality: \( x^2 - 3x - 6 - 2x < 0 \)
02
- Simplify the inequality
Combine the like terms: \( x^2 - 5x - 6 < 0 \)
03
- Factor the quadratic expression
Factor the quadratic expression on the left side: \( (x - 6)(x + 1) < 0 \)
04
- Find the critical points
Set each factor equal to zero to find the critical points: \( x - 6 = 0 \rightarrow x = 6 \)\( x + 1 = 0 \rightarrow x = -1 \)
05
- Test intervals around critical points
Test the intervals between and outside the critical points to determine where the expression is negative. The critical points divide the number line into three intervals: 1. \( (-\rightarrow, -1) \) 2. \( (-1, 6) \) 3. \( (6, \rightarrow) \)Choose test points from each interval (e.g., \( x = -2, 0, 7 \)) and plug them into \( (x - 6)(x + 1) \).
06
- Determine sign of each interval
Evaluate each test point in their respective intervals:1. For \( x = -2 \): \( (-2 - 6)(-2 + 1) = 8 \) (positive)2. For \( x = 0 \): \( (0 - 6)(0 + 1) = -6 \) (negative)3. For \( x = 7 \): \( (7 - 6)(7 + 1) = 8 \) (positive)
07
- Write the solution in interval notation
Since we need the expression to be less than 0, we select the interval where the product is negative, which is \( -1 < x < 6 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inequality Solving
Solving inequalities is similar to solving equations, with one key difference: we are looking for a range of values that satisfy the inequality, not just a single value. To solve a quadratic inequality, we can follow a few systematic steps:
1. Move all terms to one side of the inequality to set it to zero. This helps in forming a quadratic expression.
2. Factor the quadratic expression if possible. Factoring helps us identify the points that divide the number line into regions where the sign of the expression might change.
3. Identify the critical points by setting each factor equal to zero.
4. Test intervals between and outside the critical points. Plug in test points from each region into the original inequality to determine if the expression is positive or negative.
5. Select the intervals that satisfy the inequality and write the solution in interval notation.
Let's dive deeper into each of these steps to better understand how they help solve quadratic inequalities.
1. Move all terms to one side of the inequality to set it to zero. This helps in forming a quadratic expression.
2. Factor the quadratic expression if possible. Factoring helps us identify the points that divide the number line into regions where the sign of the expression might change.
3. Identify the critical points by setting each factor equal to zero.
4. Test intervals between and outside the critical points. Plug in test points from each region into the original inequality to determine if the expression is positive or negative.
5. Select the intervals that satisfy the inequality and write the solution in interval notation.
Let's dive deeper into each of these steps to better understand how they help solve quadratic inequalities.
Factoring Quadratics
Factoring quadratics involves breaking down the quadratic expression into a product of two binomials. For example, consider the quadratic expression we have: \(x^2 - 5x - 6 < 0\).
To factor this, we look for two numbers that multiply to the constant term (-6) and add up to the coefficient of the linear term (-5). Those numbers are -6 and 1. So, we can write:
\[\begin{align*}x^2 - 5x - 6 = (x - 6)(x + 1)\end{align*}\]
Once factored, the expression \((x - 6)(x + 1) < 0\) reveals the critical points (where the expression equals zero) to be x = 6 and x = -1. These points are essential for finding where the inequality holds true.
Factoring simplifies the process of finding the intervals. Itβs crucial to practice factoring since it is a foundational skill needed for solving quadratic inequalities.
To factor this, we look for two numbers that multiply to the constant term (-6) and add up to the coefficient of the linear term (-5). Those numbers are -6 and 1. So, we can write:
\[\begin{align*}x^2 - 5x - 6 = (x - 6)(x + 1)\end{align*}\]
Once factored, the expression \((x - 6)(x + 1) < 0\) reveals the critical points (where the expression equals zero) to be x = 6 and x = -1. These points are essential for finding where the inequality holds true.
Factoring simplifies the process of finding the intervals. Itβs crucial to practice factoring since it is a foundational skill needed for solving quadratic inequalities.
Interval Notation
Interval notation is a way of writing subsets of the real number line. It is especially useful in expressing the solution to inequalities. For example, the solution to our inequality \((x - 6)(x + 1) < 0\) is found by identifying intervals around critical points (-1 and 6).
The number line is divided into three intervals: \((-\infty, -1)\), \((-1, 6)\), and \((6, \infty)\). We test points within these intervals to see where the inequality holds true. After testing, we find that the quadratic expression is negative in the interval \((-1, 6)\).
We write the solution in interval notation as \((-1, 6)\). This notation is concise and clearly communicates the range of values where the inequality is true. Remember:
The number line is divided into three intervals: \((-\infty, -1)\), \((-1, 6)\), and \((6, \infty)\). We test points within these intervals to see where the inequality holds true. After testing, we find that the quadratic expression is negative in the interval \((-1, 6)\).
We write the solution in interval notation as \((-1, 6)\). This notation is concise and clearly communicates the range of values where the inequality is true. Remember:
- Parentheses \(( )\) indicate the endpoints are not included.
- Brackets \([ ]\) indicate the endpoints are included.
Critical Points
Critical points are the x-values that make the expression equal to zero. They are pivotal in determining where the inequality changes sign. For our problem, after factoring, we found the critical points by solving \(x - 6 = 0\) and \(x + 1 = 0\), resulting in:
\[x = 6 \quad \text{and} \quad x = -1\]
These critical points divide the number line into intervals. We need to determine which of these intervals make our inequality \((x - 6)(x + 1) < 0\) true. This process involves:
If we use these steps methodically, we can pinpoint the solution efficiently.
\[x = 6 \quad \text{and} \quad x = -1\]
These critical points divide the number line into intervals. We need to determine which of these intervals make our inequality \((x - 6)(x + 1) < 0\) true. This process involves:
- Selecting test points in each interval.
- Substituting these points into the factored expression.
- Evaluating the sign of the product.
If we use these steps methodically, we can pinpoint the solution efficiently.