Solution of the Inequality

Section 2.7
Question 6



Question 6

x^2 - 2x - 8 ≥ 0

Let x = -2

(-2)^2 - 2(-2) - 8 ≥ 0

4 + 4 - 8 ≥ 0

8 - 8 ≥ 0

0 ≥ 0....true statement

Let x = 0

(0)^2 -2(0) - 8 ≥ 0

0 - 8 ≥ 0

- 8 ≥ 0...false statement

Let x = -4

(-4)^2 - 2(-4) - 8 ≥ 0

16 + 8 - 8 ≥ 0

16 ≥ 0...true statement

Let x = 1

(1)^2 - 2(1) - 8 ≥ 0

1 - 2 - 8 ≥ 0

1 - 10 ≥ 0

-9 ≥ 0...false statement

There are two solutions to the inequality:

x = -2 & x = -4

If this is right, the rest are done the same way.

Yes?

 

correct

 

Section 2.7 does not look too bad.

 

Area of Rectangle As Function of x