Posted by **Mota** on Friday, January 22, 2010 at 7:49am.

find the exact values when :

x^2+2x >= 5

- math -
**bobpursley**, Friday, January 22, 2010 at 8:57am
x^2+2x-5=>0

Well, the issue is what are the factors on the left.

Using the quadratic equation

x=(-2+- sqrt (4+4*5))/2=-1+-sqrt6

(x+1-sqrt6)(x+1+sqrt6)>=0

Well, for the left side to be > 0, both factors have to be positive or negative.

Take x=-100000. That works. So negative x have to be considered.

Case I

x+1-sqrt 6<0 And x+1+sqrt6<0

x<-1+sqrt6 AND x<=-1-sqrt6

Since it has to be AND

then x<=-1-sqrt6

Now, when are both positive:

Case II

x+1-sqrt6>=0 AND x+1-sqrt6>=0

x>-1+sqrt6 AND x>-1+sqrt6

Since that is AND, then

x>-1+sqrt6

So finally, the exact values are

-1+sqrt6<x OR

x<-1-sqrt6

Check my logic.

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