What is (x^2-1)*sqrt(x^2-x-2)>= 0 ?

The solution to (x^2-1)*sqrt(x^2-x-2)>= 0 is x = 2

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(x^2-1)*sqrt(x^2-x-2)>= 0

(x^{2} - 1) \cdot x^{2} - x - 2 \geq 0

x \leq - 1 or x \geq 2

Interval Notation

(- \infty, - 1] \cup [2, \infty)

Solution steps

(x^{2} - 1) x^{2} - x - 2 \geq 0

Identify the intervals

x \leq - 1 or x \geq 2

\frac{( 1 - 21 - 4 x - x ^{2} )}{( x + 1 )} \geq 0

x - 1 - ∣ x ∣ \leq 0

3 - x \leq 1 - x

x + 3 + x + 2 - 2 x + 4 > 0

lo g_{x} (x) < 2

What is (x^2-1)*sqrt(x^2-x-2)>= 0 ?
The solution to (x^2-1)*sqrt(x^2-x-2)>= 0 is x<=-1\lor x>= 2