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 y-intercept = -2 so the vertex will be below the x-axis and, since the parabola opens up, there will be 2 x-intercepts/zeros.
The part of the parabola that is below the x-axis 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.)
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