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

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(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 (-)

then,

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.

Solution:

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>0

x+3-3>-3

x+0>-3

x>-3, which is the interval (-3, inf.)

x+4>0

x+4-4>-4

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<0

x<-3, which is the interval (-inf, -3)

and

x+4<0

x<-4, which is the interval (-inf, -4).

The common interval which satisfies the inequality is (-inf, -4).