Chapter 18: Chapter 18

Problem 540

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Solve \(\mathrm{X}^{2}-5 \mathrm{x}+4 \leq 0\)

Short Answer

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The solution to the inequality \(X^2 - 5X + 4 \leq 0\) is \(X \in (-\infty, 1] \cup [1, 4]\).

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TABLE OF CONTENTS

Step 1: Factor the quadratic expression

To factor the quadratic expression \(X^2 - 5X + 4\), we look for two numbers whose product is equal to the product of the constant term (4) and the coefficient of the squared term (1), and whose sum is equal to the coefficient of the linear term (-5). We find that these two numbers are -4 and -1, since \(-4 \times -1 = 4\) and \(-4 + (-1) = -5\). So, we can rewrite the expression as \((X - 4)(X - 1)\).

Step 2: Identify critical points (roots) of the expression

To find the critical points, we set each factor in the factored expression to zero and solve for \(X\): 1. \(X - 4 = 0\) -> \(X = 4\) 2. \(X - 1 = 0\) -> \(X = 1\) These critical points, \(X = 1\) and \(X = 4\), divide the number line into three test intervals: - Interval 1: \(X < 1\) - Interval 2: \(1 < X < 4\) - Interval 3: \(X > 4\)

Step 3: Use the test intervals method to determine the solution intervals

To determine the solution intervals, we test a number in each interval to see if it satisfies the inequality \((X - 4)(X - 1) \leq 0\). 1. Interval 1 (X < 1): choose \(X = 0\) \((0 - 4)(0 - 1) = (4)(-1) = -4\), which is less than zero. 2. Interval 2 (1 < X < 4): choose \(X = 2\) \((2 - 4)(2 - 1) = (-2)(1) = -2\), which is less than zero. 3. Interval 3 (X > 4): choose \(X = 5\) \((5 - 4)(5 - 1) = (1)(4) = 4\), which is greater than zero. We find that the inequality holds for intervals 1 and 2, so the solution to the inequality is given by \(X \leq 1\) and \(1 \leq X \leq 4\). In interval notation, this can be written as: \[X \in (-\infty, 1] \cup [1, 4].\]

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Chapter 18: Chapter 18

Solve \(\mathrm{X}^{2}-5 \mathrm{x}+4 \leq 0\)

Short Answer

Step by step solution

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Step 1: Factor the quadratic expression

Step 2: Identify critical points (roots) of the expression

Step 3: Use the test intervals method to determine the solution intervals

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Chapter 2

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Chapter 6

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