solve the inequality x^2 + 12x > 0 and state the solution set?

Question

i thought i knew it but my study partner said i was way off. please help!

Answer 1

Sorry, but the other answerers are wrong.

x² + 12x > 0

Factors into:

x(x + 12) > 0

So you have two numbers, x and x + 12, and you know that their product is greater than zero (i.e. positive).

That can only happen if they're BOTH positive OR they're BOTH negative.

If they're both positive, then:

x > 0 AND x + 12 > 0

which becomes:

x > 0 AND x > -12

The only way a number can be greater than both 0 and -12 is if it's greater than 0. (Think about it...-5 is greater than -12, but it's not greater than 0, right? but 3 is greater than both.) So part of the answer is x > 0.

Then, the other possibility is that both numbers are negative:

x < 0 AND x + 12 < 0

which becomes:

x < 0 AND x < -12

This time, the only way you can get both things to happen at once is for x < -12.

So the answer is:

x > 0 or x < -12

This is sometimes written as:

x ∈ (-∞, -12) ∪ (0, ∞)

- - - - - - - - - - - - - - - - - - - - - - - - -

Just to check, try some numbers.

If x = 3, then it should work:

3² + 12(3) = 9 + 36 = 45 > 0, check!

If x = -1 then it should NOT work:

(-1)² + 12(-1) = 1 - 12 = -11 which is less than 0, check!

If x = -13 then it should work:

(-13)² + 12(-13) = 169 - 156 = 13 > 0, check!

You have to do this kind of thing every time with these problems.

Answer 2

i'm a splash perplexed on the wording of your question so listed right here are the two considered one of my takes on your project: if it says a/a+2 >0 then: a+2(a/a+2)> 0 (a+2)= A>0 or even if it rather is a+2 > 0 a+2-2> 0-2= a>-2 wish i helped you out!!

Answer 3

x^2 + 12x > 0
subtract x ^ 2 from each side
12x > -x^2
divide each side by x
12 > -x
x > -12

Answer 4

x(x+12)>0
x>0
x+12>0
-12 -12
x>-12

the answer is (both): x>0 or x>-12

=)

Answer 5

x2 +12x>o
12x> -x2
12>-x2/x
12> -x
-12 > x


: D