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**angypangy** $\displaystyle x^2 + 2x - 8 > 0$

$\displaystyle (x + 1)^2 - 9 > 0$

$\displaystyle (x + 1)^2 > 9$

$\displaystyle x + 1 > 3$ But what to do about the -3. Eg Do I say:

$\displaystyle x + 1 > -3$ or $\displaystyle x + 1 < -3$

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

But then what if problem is $\displaystyle x^2 + 2x - 8 < 0$

$\displaystyle (x + 1)^2 < 9$

$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 < -3$ which is wrong. Should be

$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 > -3$.

-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???