Posted by Blair on Sunday, February 17, 2013 at 5:16pm.
1. x^2 + 5x - 6 > 0
(x+6)(x-1) > 0
critical values are x = -6 and x = 1
so the parabola y = x^2 + 5x - 6 crosses the x-axis at -6 and 1, and knowing about the basic shape, (it opens upwards), we can state that it will be above for
x < -6 OR x > 1
2. 6x^2 - 5x + 1 < 0
(6x - 1)(x + 1) < 0
critical values are .....
since we want to be below the x-axis ( < 0)
.......
take it from here, and use the notation that was taught to you,
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