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Magazine Chapter 3: Problem 94
Solve the inequality \(x^{2}-4 x+3 \geq 0 .\) Write the solution set using interval notation.
Short Answer
Expert verified
The solution set is \((-\infty, 1] \cup [3, +\infty)\).
Step by step solution
01
Find the roots of the quadratic equation
Solve the equation \( x^2 - 4x + 3 = 0 \) to find the roots. This can be done by factoring the quadratic expression. The factored form is \( (x-1)(x-3) = 0 \). Hence, the roots are \( x = 1 \) and \( x = 3 \).
02
Determine the intervals to test
The roots divide the number line into three intervals: \((-\infty, 1)\), \((1, 3)\), and \((3, +\infty)\). We need to determine on which of these intervals the inequality \(x^2 - 4x + 3 \geq 0\) holds true.
03
Test points in each interval
Choose test points from each interval to determine where the quadratic expression is non-negative. For \((-\infty, 1)\), choose \(x = 0\): \(0^2 - 4 \cdot 0 + 3 = 3 \geq 0\), so this interval satisfies the inequality. For \((1, 3)\), choose \(x = 2\): \(2^2 - 4 \cdot 2 + 3 = -1 \< 0\), so this interval does not satisfy the inequality. For \( (3, +\infty)\), choose \(x = 4\): \(4^2 - 4 \cdot 4 + 3 = 7 \geq 0\), so this interval satisfies the inequality.
04
Verify the boundary points
Check if the roots \(x = 1\) and \(x = 3\) themselves satisfy the inequality \(x^2 - 4x + 3 \geq 0\). Substitute \(x = 1 \): \(1^2 - 4 \cdot 1 + 3 = 0 \geq 0\). Substitute \(x = 3\): \(3^2 - 4 \cdot 3 + 3 = 0 \geq 0\). So, both \(x = 1\) and \(x = 3\) are included in the solution set.
05
Combine the intervals
Combine the intervals and boundary points where the inequality holds. Therefore, the solution set is \((-\infty, 1] \cup [3, +\infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
Factoring quadratics is a key concept when solving quadratic inequalities. To factor a quadratic like \(x^2 - 4x + 3 = 0\), we look for two numbers that multiply to give the constant term (3) and add to give the coefficient of the linear term (-4). Here, the factors are \( (x-1)(x-3) \). This means the quadratic can be written as \( (x-1)(x-3) = 0 \), and we can solve for the roots \( x = 1 \) and \( x = 3 \).
Factoring helps us find the critical points that break the quadratic into intervals for further testing.
Factoring helps us find the critical points that break the quadratic into intervals for further testing.
Interval Notation
Interval notation is a way of writing subsets of the real numbers. It expresses the solution set of inequalities concisely. In interval notation, a parenthesis \(()\) means that an endpoint is not included, while a bracket \([])\) means it is included.
For instance, the solution \((-fty, 1] \cup [3, +fty)\) includes all numbers less than or equal to 1 and then starts again at 3, including 3, up to infinity. This is derived from the intervals tested and their respective boundary points.
For instance, the solution \((-fty, 1] \cup [3, +fty)\) includes all numbers less than or equal to 1 and then starts again at 3, including 3, up to infinity. This is derived from the intervals tested and their respective boundary points.
Solving Inequalities
To solve quadratic inequalities, like \(x^2 - 4x + 3 \geq 0\), start by finding the roots of the associated quadratic equation \(x^2 - 4x + 3 = 0\). The roots divide the number line into intervals. Then, determine on which intervals the inequality holds true.
This involves evaluating the quadratic at points within these intervals and including the boundary points if the inequality is non-strict (\(\geq\) or \(\leq\)).
This involves evaluating the quadratic at points within these intervals and including the boundary points if the inequality is non-strict (\(\geq\) or \(\leq\)).
Testing Intervals
Testing intervals helps to identify where the quadratic inequality holds. After factoring the quadratic, the roots (\(x = 1\) and \(x = 3\)) divide the number line into three intervals: \((-fty, 1)\), \((1, 3)\), and \((3, +fty)\).
Choose a test point from each interval and substitute it into the inequality. For example:
Choose a test point from each interval and substitute it into the inequality. For example:
- For \((-fty, 1)\), test \(x = 0\). Since \(0^2 - 4 \cdot 0 + 3 = 3 \ge 0\), the interval satisfies the inequality.
- For \((1, 3)\), test \(x = 2\). Since \(2^2 - 4 \cdot 2 + 3 = -1 \(\leq\)< 0\), the interval does not satisfy the inequality.
- For \((3, +fty)\), test \(x = 4\). Since \(4^2 - 4 \cdot 4 + 3 = 7 \geq 0\), the interval satisfies the inequality.
