Solve \[\frac{|x^2 - 81|}{x^2 - 36x} < 0\]

Evaluate

\frac{x ^{2} - 81}{x ^{2} - 36 x} < 0

Find the domain

More Steps

\frac{x ^{2} - 81}{x ^{2} - 36 x} < 0, x \in (- \infty, 0) \cup (0, 36) \cup (36, + \infty)

Separate the inequality into 2 possible cases

{x^{2} - 81 > 0 x^{2} - 36 x < 0 {x^{2} - 81 < 0 x^{2} - 36 x > 0

Solve the inequality

More Steps

{x \in (- \infty, - 9) \cup (- 9, 9) \cup (9, + \infty) x^{2} - 36 x < 0 {x^{2} - 81 < 0 x^{2} - 36 x > 0

Solve the inequality

More Steps

{x \in (- \infty, - 9) \cup (- 9, 9) \cup (9, + \infty) 0 < x < 36 {x^{2} - 81 < 0 x^{2} - 36 x > 0

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

{x \in (- \infty, - 9) \cup (- 9, 9) \cup (9, + \infty) 0 < x < 36 {x \in / R x^{2} - 36 x > 0

Solve the inequality

More Steps

{x \in (- \infty, - 9) \cup (- 9, 9) \cup (9, + \infty) 0 < x < 36 {x \in / R x \in (- \infty, 0) \cup (36, + \infty)

Find the intersection

x \in (0, 9) \cup (9, 36) {x \in / R x \in (- \infty, 0) \cup (36, + \infty)

Find the intersection

x \in (0, 9) \cup (9, 36) x \in / R

Find the union

x \in (0, 9) \cup (9, 36)

Check if the solution is in the defined range

x \in (0, 9) \cup (9, 36), x \in (- \infty, 0) \cup (0, 36) \cup (36, + \infty)

Solution

x \in (0, 9) \cup (9, 36)