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+2x5=>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+1sqrt6)(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+1sqrt 6<0 And x+1+sqrt6<0
x<1+sqrt6 AND x<=1sqrt6
Since it has to be AND
then x<=1sqrt6
Now, when are both positive:
Case II
x+1sqrt6>=0 AND x+1sqrt6>=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<1sqrt6
Check my logic.
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