Question

final exam
100 points possible answered: 5/22
question 7
solve the following rational inequality \\(\frac{x - 2}{x^2 - 25} > 0\\).
state your answer using interval notation. use u for union and oo for \\(\infty\\).
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Explanation:

Step1: Factor the denominator

First, factor the denominator \(x^2 - 25\) using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\). So, \(x^2 - 25=(x - 5)(x + 5)\). The inequality becomes \(\frac{x - 2}{(x - 5)(x + 5)}>0\).

Step2: Find critical points

The critical points are the values of \(x\) that make the numerator or the denominator equal to zero.

• For the numerator \(x - 2 = 0\), we get \(x = 2\).
• For the denominator \((x - 5)(x + 5)=0\), we get \(x = 5\) and \(x=- 5\). These points divide the real number line into intervals: \((-\infty,-5)\), \((-5,2)\), \((2,5)\), and \((5,\infty)\).

Step3: Test intervals

We test a value from each interval in the inequality \(\frac{x - 2}{(x - 5)(x + 5)}>0\):

• For the interval \((-\infty,-5)\), let's choose \(x=-6\). Then \(\frac{-6 - 2}{(-6 - 5)(-6 + 5)}=\frac{-8}{(-11)(-1)}=\frac{-8}{11}<0\), so this interval does not satisfy the inequality.
• For the interval \((-5,2)\), let's choose \(x = 0\). Then \(\frac{0 - 2}{(0 - 5)(0 + 5)}=\frac{-2}{(-5)(5)}=\frac{-2}{-25}=\frac{2}{25}>0\), so this interval satisfies the inequality.
• For the interval \((2,5)\), let's choose \(x = 3\). Then \(\frac{3 - 2}{(3 - 5)(3 + 5)}=\frac{1}{(-2)(8)}=\frac{1}{-16}<0\), so this interval does not satisfy the inequality.
• For the interval \((5,\infty)\), let's choose \(x = 6\). Then \(\frac{6 - 2}{(6 - 5)(6 + 5)}=\frac{4}{(1)(11)}=\frac{4}{11}>0\), so this interval satisfies the inequality.

We also need to note that \(x=-5\) and \(x = 5\) make the denominator zero, so they are not included in the solution set, and \(x = 2\) makes the numerator zero, and at \(x = 2\) the expression is equal to zero, so it is not included as we need the expression to be greater than zero.

Answer:

\((-5,2)\cup(5,\infty)\)