Question 505829: 4x^2-5 < 8x
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! 4x^2-5 < 8x
4x^2-8x-5<0
4x^2-8x-5=0 Change equation to equal to zero temporarily.
4*-5=-20, two factors of -20 whose sum is -8 are -10 and +2.
4x^2+2x-10x-5=0
2x(2x+1)-5(2x+1)=0
(2x-5)(2x+1)=0
x=5/2, x=-1/2 These are the x intercepts on the graph. We have to test the equation on the second line above and below the intercepts to find where y<0
x=-1:y=4+8-5=+7 positive
x=0:y=0+0-5=-5 negative
x=3:y=36-24-5=+7 positive
So y is negative between -1/2 and 5/2.
-1/2
.
Ed
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