Which real numbers satisfies the inequality (x+3)(x+4)>0?

Expert Answers
hala718 eNotes educator| Certified Educator


>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)

neela | Student

(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)


giorgiana1976 | Student

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).