Sorry for the horrible handwriting and drawings.

http://picpaste.com/Question-zdfK0hLF.png

NOTE: It deletes itself in 6 hours from posing

EDIT:

If ab<0

then,

a<0 b>0

OR

a>0 b<0

http://picpaste.com/Question-zdfK0hLF.png

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

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.

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: |
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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: |
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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: |
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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.

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