math

find the exact values when :
x^2+2x >= 5

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  1. 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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    bobpursley

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