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Chapter 2: Problem 5

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$x^{2}-5 x+4>0$$

Short Answer

Expert verified
The solution to the inequality \(x^{2}-5 x+4>0\) is \((-\infty, 1] \cup (4, +\infty)\)

Step by step solution

01

Find critical values

To find the critical numbers, set the equation \(x^{2}-5 x+4=0\). This can be factored to get \((x-4)(x-1) = 0\), which gives the solution \(x=1, x=4\). These are the critical values, because they are where the equation equals zero and potentially changes sign.
02

Graph and test intervals

On a number line, plot the critical numbers \(1\) and \(4\). These numbers divide the line into the intervals \( (-\infty, 1) \), \( (1, 4) \), and \( (4, +\infty) \). Test a number from each interval in the original inequality \(x^{2}-5 x+4>0\). For instance, for the interval \((-\infty, 1)\), let's take \(x=0\); for \((1, 4)\), take \(x=2\); and for \((4, +\infty)\), take \(x=5\). We then find that for \(x=0\), \(0 - 0 + 4 > 0\), which is true, so the interval \((-\infty, 1)\) is a solution set. For \(x=2\), \(2^2 - 5*2 + 4 > 0\) is false, so the interval \((1, 4)\) is not a solution. For \(x=5\), \(5^2 - 5*5 + 4 > 0\) is true, so the interval \((4, +\infty)\) is a solution set.
03

Write the solution in interval notation

Express the solution set in interval notation. The solution is \((-\infty, 1] \cup (4, +\infty)\). This includes all real numbers less than or equal to 1, and all real numbers greater than 4.

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