Help me solve this inequality x^2-x-2>_ 0?

Question

the inequality is supposed to be greater than or equal to

Answer 1

x^2-x-2>0

(x+1)(x-2)>0

x+1>0, x-2>0

so, the solution for this inequality is:

x>-1, x>2

Answer 2

x^2 - x - 2 >= 0

(x + 1)(x - 2) >= 0

Let's split into 3 sections:

If x<= -1

x + 1 <= -1 + 1 = 0

x - 2 <= -1 - 2 = -3 <0

Thus (x + 1)(x - 2) >= 0 (a product of two negative numbers is positive)

If -1 < x < 2

x + 1 > -1 + 1 = 0

x - 2 < 2 - 2 = 0

Thus (x + 1)(x - 2) < 0 (a product of a positive number and a negative number is negative).

If x >= 2

x + 1 > 2 + 1 = 3 > 0

x - 2 > 2 - 2 = 0

Thus, (x + 1)(x - 2) >= 0 (a product of two positive numbers is positive)

x <= -1 or x >= 2

Answer 3

x^2-x-2>0

this is quadratic function, the graph of this function is a parable directed upward cutting the axis of X in points:

x^2-x-2=0

x= -1

x=2

the sign of inequality is >0, so we take the external values:

the solution for this inequality is:

x< -1 v x>2

Answer 4

x² - x - 2 >= 0

x² - x >= 2

x² - 1/2x = 2 + (- 1/2)²

x² - 1/2x = 8/2 + 1/4

(x - 1/2x)² = 9/4

x - 1/2 = 3/2

1st factor:

= x - 1/2 - 3/2

= x - 2

2nd factor:

= x - 1/2 + 3/2

= x + 1

Answer: x = 2, x = - 1