Solve 0 > x^2 + 5x 2.
Let's think about the graph of y = x^2 + 5x  2.
Since the leading coefficient is positive the parabola opens up.
The yintercept = 2 so the vertex will be below the xaxis and, since the parabola opens up, there will be 2 xintercepts/zeros.
The part of the parabola that is below the xaxis is where y < 0, and that part is between the zeros.
So let's find the zeros of y(x) using the quadratic formula:
 b ± sqrt(b^2  4ac)  5 ± sqrt(5^2  4(1)(2))  5 ± sqrt(25 + 8)
x =  =  = 
2a 2(1) 2
 5 ± sqrt(33)
x =  => answer is A.
2
(The zeros are not in the interval because y < 0.)
12/15/2013

Steve S.
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