Problem 12
Question
What whole numbers can replace \(x\) so that the following statement is true?
Step-by-Step Solution
Verified Answer
Answer: 2
1Step 1: Find the critical points
To find the critical points, set the quadratic expression to zero and solve for x:
\(x^2 - 4x + 3 = 0\)
We can factor the equation as follows:
\((x-3)(x-1) = 0\)
Thus, the critical points are \(x = 1\) and \(x = 3\).
2Step 2: Create a sign chart
We create a sign chart to determine the intervals where the inequality \(x^2 - 4x + 3 < 0\) is satisfied.
Divide the real number line into intervals using the critical points:
- Interval 1: \(x < 1\)
- Interval 2: \(1 < x < 3\)
- Interval 3: \(x > 3\)
Now, we'll test each interval to see whether the inequality is true.
3Step 3: Test each interval
For Interval 1 (\(x < 1\)), choose a test point \(x = 0\):
\((0-3)(0-1) = 3 \cdot 1 = 3 > 0\)
The inequality is not true for this interval.
For Interval 2 (\(1 < x < 3\)), choose a test point \(x = 2\):
\((2-3)(2-1) = -1 \cdot 1 = -1 < 0\)
The inequality is true for this interval.
For Interval 3 (\(x > 3\)), choose a test point \(x = 4\):
\((4-3)(4-1) = 1 \cdot 3 = 3 > 0\)
The inequality is not true for this interval.
Since the inequality is true for Interval 2 (\(1 < x < 3\)), we can conclude that the whole numbers that satisfy the inequality are \(2\).
Key Concepts
Factoring Quadratic EquationsSign Chart MethodCritical Points in InequalitiesTest Point Method
Factoring Quadratic Equations
In order to solve quadratic inequalities, like determining the value of whole numbers that make the statement \(x^2 - 4x + 3 < 0\) true, we often start by factoring quadratic equations. This process involves writing the quadratic expression in the form of a product of two simpler linear factors. For example:\(
\) To factor the equation \(x^2 - 4x + 3 = 0\), we search for two numbers that multiply to give the constant term (\(3\)) and add up to the coefficient of \(x\), which is \( -4\). These numbers are \( -1\) and \( -3\), leading us to the factored form \( (x - 1)(x - 3) = 0 \). By doing this, we can easily identify the critical points - where the expression equals zero - which are crucial for defining intervals in the sign chart method.
Understanding how to factor quadratic equations efficiently is fundamental for solving inequalities because it simplifies the problem and sets the stage for further analysis through the sign chart method.
\) To factor the equation \(x^2 - 4x + 3 = 0\), we search for two numbers that multiply to give the constant term (\(3\)) and add up to the coefficient of \(x\), which is \( -4\). These numbers are \( -1\) and \( -3\), leading us to the factored form \( (x - 1)(x - 3) = 0 \). By doing this, we can easily identify the critical points - where the expression equals zero - which are crucial for defining intervals in the sign chart method.
Understanding how to factor quadratic equations efficiently is fundamental for solving inequalities because it simplifies the problem and sets the stage for further analysis through the sign chart method.
Sign Chart Method
The sign chart method is a visual aid that helps in solving quadratic inequalities. Once you have determined the critical points by factoring the quadratic equation, you can use these points to set up a sign chart. This involves dividing the real number line into intervals based on the critical points. Here's how you'd set up a sign chart for our initial problem:\(
\) With critical points at \(x = 1\) and \(x = 3\), we divide the number line into three intervals: \(x < 1\), \(1 < x < 3\), and \(x > 3\). These intervals help us to test where the inequality expression changes its sign - from positive to negative or vice versa. The goal is to determine on which intervals the original inequality, such as \(x^2 - 4x + 3 < 0\), is satisfied. By plotting these intervals on a sign chart, we can visually inspect and predict the behavior of the quadratic expression within each interval without solving the inequality at every single point.
\) With critical points at \(x = 1\) and \(x = 3\), we divide the number line into three intervals: \(x < 1\), \(1 < x < 3\), and \(x > 3\). These intervals help us to test where the inequality expression changes its sign - from positive to negative or vice versa. The goal is to determine on which intervals the original inequality, such as \(x^2 - 4x + 3 < 0\), is satisfied. By plotting these intervals on a sign chart, we can visually inspect and predict the behavior of the quadratic expression within each interval without solving the inequality at every single point.
Critical Points in Inequalities
Critical points in inequalities are the values of the variable that make the quadratic expression equal to zero. They play a pivotal role in determining the solution set of the inequality. For the given quadratic inequality \(x^2 - 4x + 3 < 0\), we found the critical points to be \(x = 1\) and \(x = 3\) after factoring the expression. These critical points split the entire range of possible values into distinct intervals where the signs of the values taken by the quadratic expression might change. It is at and around these critical points that we focus our attention to determine where the inequality holds true. Understanding the concept of critical points is essential because it helps predict the intervals that need closer examination using the test point method for confirming where the inequality is satisfied.
Test Point Method
Once we establish intervals using the critical points, we employ the test point method to find out which intervals satisfy the inequality. The test point method involves selecting a value from each interval, which we call the 'test point', and substituting it into the inequality. Through this method, we determine the 'sign' of each interval. For the inequality \(x^2 - 4x + 3 < 0\), we selected test points \(0\), \(2\), and \(4\) for the corresponding intervals \(x < 1\), \(1 < x < 3\), and \(x > 3\) respectively.
By calculating \(f(0)\), \(f(2)\), and \(f(4)\), where \(f(x)\) is our quadratic expression, we noted the sign of the expression within each interval. Since the expression is only negative in the interval \(1 < x < 3\) as indicated by the test point \(x = 2\), we can confidently say that the only whole number satisfying the inequality is \(2\). This method provides a clear-cut way of verifying which intervals fulfill the inequality without extensive algebraic manipulations, saving time and simplifying the process substantially for students.
By calculating \(f(0)\), \(f(2)\), and \(f(4)\), where \(f(x)\) is our quadratic expression, we noted the sign of the expression within each interval. Since the expression is only negative in the interval \(1 < x < 3\) as indicated by the test point \(x = 2\), we can confidently say that the only whole number satisfying the inequality is \(2\). This method provides a clear-cut way of verifying which intervals fulfill the inequality without extensive algebraic manipulations, saving time and simplifying the process substantially for students.
Other exercises in this chapter
Problem 12
Make use of either or both the power rule for products and the power rule for powers to simplify each expression. $$ \left(10^{6} \cdot 10^{12} \cdot 10^{5}\rig
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For the following problems, use the order of operations to find each value. $$8 \cdot 4 \div 16+5$$
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For the following problems, simplify the expressions. $$ \frac{8 \cdot 6}{2}+\frac{9 \cdot 9}{3}-\frac{10 \cdot 4}{5} $$
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