Solve \frac{x^4+x^2+1}{x^2-4x-5}<0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

- 1 < x < 5

Alternative Form

x \in (- 1, 5)

Evaluate

\frac{x ^{4} + x ^{2} + 1}{x ^{2} - 4 x - 5} < 0

Find the domain

More Steps

\frac{x ^{4} + x ^{2} + 1}{x ^{2} - 4 x - 5} < 0, x \in (- \infty, - 1) \cup (- 1, 5) \cup (5, + \infty)

Set the numerator and denominator of \frac{x ^{4} + x ^{2} + 1}{x ^{2} - 4 x - 5} equal to 0 to find the values of x where sign changes may occur

x^{4} + x^{2} + 1 = 0 x^{2} - 4 x - 5 = 0

Since the left-hand side is always positive,and the right-hand side is always 0,the statement is false for any value of x

x \in / R x^{2} - 4 x - 5 = 0

Calculate

More Steps

x \in / R x = 5 x = - 1

Determine the test intervals using the critical values

x < - 1 - 1 < x < 5 x > 5

Choose a value form each interval

x_{1} = - 2 x_{2} = 2 x_{3} = 6

To determine if x < - 1 is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < - 1 is not a solution x_{2} = 2 x_{3} = 6

To determine if - 1 < x < 5 is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < - 1 is not a solution - 1 < x < 5 is the solution x_{3} = 6

To determine if x > 5 is the solution to the inequality,test if the chosen value x = 6 satisfies the initial inequality

More Steps

x < - 1 is not a solution - 1 < x < 5 is the solution x > 5 is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is - 1 < x < 5

- 1 < x < 5

Check if the solution is in the defined range

- 1 < x < 5, x \in (- \infty, - 1) \cup (- 1, 5) \cup (5, + \infty)

Solution

- 1 < x < 5

Alternative Form

x \in (- 1, 5)

Show Solution