How to solve x^2 + 2x - 8 > 0 algebraically

But what to do about the -3. Eg Do I say:

or

It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

But then what if problem is

and which is wrong. Should be

and .

-4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

So what is the rule for bit when get square root of either side???

Re: How to solve x^2 + 2x - 8 > 0 algebraically

Quote:

Originally Posted by

**angypangy** But what to do about the -3. Eg Do I say:

or

It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

But then what if problem is

and

which is wrong. Should be

and

.

-4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

So what is the rule for bit when get square root of either side???

You need to say that

Then either

Re: How to solve x^2 + 2x - 8 > 0 algebraically