drakehikcs
Which of the following describes all the values of x such that 3x2 − 5x − 12 > 0?
-4/3 < x < 3
x > 3
x <
x > 3 or x < −4
x > 3 or x <-4/3
3x2 − 5x − 12 = (x − 3)(3x + 4). So the inequality 3x2 − 5x − 12 > 0 can be rewritten (x − 3)(3x + 4) > 0. For this equation, why do you test for both positive and negative values?
I understand x>3, and x>-4/3, which is the single requirement that x>3, but then considering the possibility that x − 3 < 0 and 3x + 4 < 0. If x − 3 < 0, is that general practice? Can you help me understand this?
We have: 3x^2 − 5x − 12 > 0
=> (x − 3)(3x + 4) > 0
Consider the function: y = (x − 3)(3x + 4)
The roots of y = 0 are: x = 3 or x = -4/3
The graph of this is presented below:
Attachment:
1.JPG [ 19.25 KiB | Viewed 3860 times ]
Since we need (x − 3)(3x + 4) > 0, i.e. y > 0, we look at the positive part of y
The y-values are positive when x > 3 or x < -4/3 - hence, this is the solution
Alternative approach:
Case 1: Positive * Positive > 0
Thus: x - 3 > 0 and x + 4/3 > 0 => x > 3 and x > -4/3 => x > 3 (common region)
Case 2: Negative * Negative > 0
Thus: x - 3 < 0 and x + 4/3 < 0 => x < 3 and x < -4/3 => x < -4/3 (common region)
Thus, the solution: x > 3 or x < -4/3
Hope this helps!