### Maths Question

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NOTE: It deletes itself in 6 hours from posing

EDIT:
If ab<0
then,
a<0 b>0
OR
a>0 b<0
Last edited on
What's the question?
Your problem should can be solved like this.
Solve:
(x+2)(x-2)<0.
Therefore (x-2)<0 and (x+2)>0 or (x-2)>0 and (x+2)<0. The latter is not possible because (x-2)>0 implies (x+2)>0. Therefore the solution to the problem is -2<x<2.
Last edited on
tition wrote:
The latter is not possible because (x-2)>0 implies (x+2)>0.

I do not understand how x-2>0 implies that x+2>0, in fact as far as I know it implies the opposite.

@kbw please click on the (more than likely) blue underlined part of my first post.
Script Coder wrote:
tition wrote:
The latter is not possible because (x-2)>0 implies (x+2)>0.

I do not understand how x-2>0 implies that x+2>0, in fact as far as I know it implies the opposite.

x-2 > 0 means that x > 2.

Therefore, x+2 > 2+2 > 0.
Script Coder wrote:
I do not understand how x-2>0 implies that x+2>0, in fact as far as I know it implies the opposite.

x - 2 > 0
x > 2

x + 2 > 0
x > -2

In English, if x is greater than two, than x has to be greater than -2.
I understand, thanks guys.
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