How do you do x^2+2X-8<0 ??

Question

Ok, so I know that you factor it so it would be:
(x+4)(x-2)<0,
and you find the critical values, so it would be:
x=-4 and x=2

What do you do after that (what would the solution set be) and how do you graph the solution set?

Answer 1

ok now draw a number line with x=-4 and x=2 on it.

you have to test 5 regions out.
the region less than -4
-4
the region between -4 and 2
2
the region above 2


the region less than -4: substitute -5 into (x+4)(x-2)<0,
(-5+4)(-5-2)<0, this is false so we don't want this.

-4:substitute -4 into (x+4)(x-2)<0, not true

the region between -4 and 2 :substitute 1 into (1+4)(1-2)<0, true so this is a region we want to consider but doesnt not include -4

the region above 2: substitute 3 into (3+4)(3-2)<0, false


so you would say that (-4,2)
the round brackets tell us that we go right up to -4 and 2 but do not include them- put this info on a number line open circle at -4 and 2 and a line connectign them

Answer 2

X does not equal -4 or 2. They would appear as open circles on your line gragh.

2 would have an circle.
You would have a solid line from 2 to -4.
-4 would have a circle.

If you try a number like -5 or 3, your answer would be greater than 0.

Answer 3

factor (x +4)(x-2) < 0
set both equal to zero
x + 4 < 0 x - 2 < 0
x < -4 x < 2

therefore
x<2

Answer 4

draw the arrow line and put -4 on theleft and 2 on the right with some space on between, circle each one but dont fill inthe circle, and color/point ur arrow to the left

Answer 5

I HAV NO IDEA WHOA