Questions: Solve the following rational inequality (x-2)/(x^2-16)>0. State your answer using interval notation. Use U for union and oo for infinity.

Solution Steps

To solve the rational inequality \(\frac{x-2}{x^{2}-16}>0\), we need to determine the values of \(x\) for which the expression is positive. This involves finding the critical points where the numerator and denominator are zero, and then testing the intervals between these points to see where the inequality holds.

1. Find the zeros of the numerator and the denominator.
2. Determine the critical points and the intervals they create on the number line.
3. Test each interval to see where the inequality is satisfied.
4. Combine the intervals where the inequality holds true.

To solve the inequality \( \frac{x-2}{x^2-16} > 0 \), we first find the critical points by setting the numerator and denominator to zero. The numerator \( x - 2 = 0 \) gives us \( x = 2 \). The denominator \( x^2 - 16 = 0 \) factors to \( (x - 4)(x + 4) = 0 \), yielding the critical points \( x = -4 \) and \( x = 4 \).

The critical points divide the number line into the following intervals:

1. \( (-\infty, -4) \)
2. \( (-4, 2) \)
3. \( (2, 4) \)
4. \( (4, \infty) \)

We will test a point from each interval to determine where the inequality holds:

• For the interval \( (-\infty, -4) \), choose \( x = -5 \): \[ \frac{-5 - 2}{(-5)^2 - 16} = \frac{-7}{25 - 16} = \frac{-7}{9} < 0 \]

• For the interval \( (-4, 2) \), choose \( x = 0 \): \[ \frac{0 - 2}{0^2 - 16} = \frac{-2}{-16} = \frac{1}{8} > 0 \]

• For the interval \( (2, 4) \), choose \( x = 3 \): \[ \frac{3 - 2}{3^2 - 16} = \frac{1}{9 - 16} = \frac{1}{-7} < 0 \]

• For the interval \( (4, \infty) \), choose \( x = 5 \): \[ \frac{5 - 2}{5^2 - 16} = \frac{3}{25 - 16} = \frac{3}{9} > 0 \]

The inequality \( \frac{x-2}{x^2-16} > 0 \) holds true in the intervals \( (-4, 2) \) and \( (4, \infty) \).

Final Answer