SOLUTION: given g(x)= -2x^2-8x+12.......what are the solutuions for g(x)=0, and g(x)<0.

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Question 17346: given g(x)= -2x^2-8x+12.......what are the solutuions for g(x)=0, and g(x)<0.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
g(x)= -2x^2-8x+12 =0 or less than zero
first solve g(x)=0 using
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-8%29+%2B-+sqrt%28+%28-8%29%5E2-4%2A%28-2%29%2A%2812%29+%29%29%2F%282%2A%28-2%29%29+
=-2%2B-sqrt%2810%29 are the 2 solutions in the case when g(x)=0
For the case g(x)<0 we have
g(x)= [x-(-2+10^0.5)][x-(-2-10^0.5)]<0..product of 2 factors will be negative if one of the 2 is negative and the other positive
hence g(x)<0 if x lies between (-2+sqrt10) and (-2-sqrt10)