Which real numbers satisfies the inequality (x+3)(x+4)>0?
>0 means that the function on the left side should be positive.
since we have 2 numbers mutliplying by each other, then either both of them are positive (+) or both are negative (-)
x+3> 0 and x+4>0
==> x>-3 and X>-4 .... then the interval would be (-3, inf)
now lets wok on the negative side
both numbers are negative (-), then
x+3<0 and x+4< 0
x<-3 and x<-4 .... the interval is (-inf, -4)
then the inequality is ture iff (-inf, -4) OR ( -3, inf)
or means either the first or the second but not both
===> (-inf, -4) U (-3, inf)
(x+3)(x+4) > 0.
To find the real numbers that satisfies the above product.
Let f(x) = (x+3)(x+4)
So f(x) vanishes when eother of the factors is zero.
Consider the sign of f(x) = (x+3)(x+4). When x<-4, both factors are negative . So the product of 2 negatives is positve. So f(x) = (x+3)(x+4) > 0 when x <-4.
When x=-4. f(x) = 0 as (x+3)(-4+4) = 0
When -4 < x<-3, f(x) has x+3 negative and x+4 is positive. So the product of a negative and positive factor is negative.
When x>-3, both factors are positive. Therefore f(x) is positive.
So for f(x) > 0, x <-4 or x >-3. That means x should be outside the interval (-3 , -4)
To satisfy the inequality, both factors of the product have to have the same sign, that means that, if (x+3) is positive, (x+4)>, too, and reverse.
Let's solve both cases.
First, let's consider both factors as being positive:
x>-3, which is the interval (-3, inf.)
x>-4, which is the interval (-4, inf.)
The common interval which satisfies the inequality is (-3, inf.).
The other case is when both factors are negative.
x<-3, which is the interval (-inf, -3)
x<-4, which is the interval (-inf, -4).
The common interval which satisfies the inequality is (-inf, -4).