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January 30, 2015

January 30, 2015

Posted by **Blair** on Sunday, February 17, 2013 at 5:16pm.

6x^2-5x+1<0

Solve (using any method), write your answer in interval notation and graph the solution set.

- MATH- Algebra 2 -
**Reiny**, Sunday, February 17, 2013 at 5:51pm1. 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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