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,
x^2-6>-5x
6x^2-5x+1<0
Solve (using any method), write your answer in interval notation and graph the solution set.
2 answers
[1/3, 1/2]