-2x^2-8x is less than 6?

Question

show work please

Answer 1

-2x^2-8x < 6

-2x^2-8x-6<0 /(-2)
x^2+4x+3>0
-4+/- root:(4^2-4*1*3) / 2

1- (-4+2)/2= -1 2- (-4-2)/2= -3

x<-3 x>-1

Answer 2

-2x² -8x < 6

Divide both sides by -2 (reverse the inequality):
x² + 4x > -3

Add 3 to both sides:
x² + 4x + 3 > 0

Factoring:
(x + 1)(x + 3) > 0

Thus the product must be greater than zero. A product is positive if you multipy two negative numbers or two positive numbers.

Two solutions are therefore:
(x + 1) > 0 and (x + 3) > 0
- or -
(x + 1) < 0 and (x + 3) < 0

In the first case, this is simplified to:
x > -1 and x > -3
This is true only when x > -1

In the second case:
x < -1 and x < -3
This is only try when x < -3

So the answer is x < -3 or x > -1

In other words, x can be any real number except those between -3 and -1, inclusive.

On a number line this would look like an open circle on -3 with an arrow to the left, and an open circle on -1 with an arrow to the right.

You can check your answers, by trying some points either side.
x = -4 (this should work)
-2(-4)² - 8(-4) =?
-2(16) + 32 =?
-32 + 32 =?
0 < 6

x = -3 (this should be exactly 6 and *not* work)
-2(-3)² - 8(-3) =?
-2(9) + 24 =?
-18 + 24 =?
6 = 6

x = -2 (this should *not* work)
-2(-2)² - 8(-2) =?
-2(4) + 16 =?
-8 + 16 =?
8 > 6

x = -1 (this should equal 6 and *not* work)
-2(-1)² - 8(-1) =?
-2(1) + 8 =?
-2 + 8 =?
6 = 6

x = 0 (this should work)
-2(0)² -8(0) =?
0 - 0 =?
0 < 6

These all confirm our answer:
x < -3 or x > -1

Answer 3

-2x^2 - 8x < 6

Move everything to one side:

2x^2 + 8x + 6 > 0

Divide by 2:

x^2 + 4x + 3 > 0

Factor:

(x+3)(x+1) > 0

For this to be the case, either x+3 > 0 and x+1 > 0, or x+3 < 0 and x+1 < 0 (since the product of two negative numbers is positive).

To have x+3>0 and x+1>0, you must have x>-3 and x>-1, so x>-1.

To have x+3<0 and x+1<0, you must have x<-3 and x<-1, so x<-3.

Therefore, your solution is x>-1 or x<-3.