6x³ > 7x² + 3x
There are three cases to consider, when
x > 0, when x = 0, and when x < 0.
--------------------------
Case 1: x > 0
6x³ > 7x² + 3x
Since x is positive, we may divide
through by x without changing the
direction of the inequality.
6x² > 7x + 3
6x² - 7x - 3 > 0
(2x-3)(3x+1) > 0
Setting each of those two parentheses =
0 gives critical numbers and
. We can ignore the negative
since x is positive for this case, and
so we take 0 as the lower critical
number. So the critical numbers are 0
and .
If x is between the two critical
valuee, say x=1, the inequality becomes
[2(1)-3][3(1)+1] > 0
(-1)(4) > 0
-4 > 0
That's false, so the inequality does
not hold in the interval between the
critical values 0 and .
If x > , say x=2, the
inequality becomes
[2(2)-3][3(2)+1] > 0
(4-3)(6+1) > 0
(1)(7) > 0
7 > 0
That's true so the inequality is
true when x is in the interval (,∞)
So for case 1, the inequality has
solution set
(,∞)
----------------------------
Case 2: x=0
This case is ruled out because
6x³ > 7x² + 3x becomes 0 > 0 which is
false.
----------------------------
Case 3: x < 0
6x³ > 7x² + 3x
Since x is negative, if we divide
through by x we must change the
direction of the inequality.
6x² < 7x + 3
6x² - 7x - 3 < 0
(2x-3)(3x+1) < 0
Setting each of those two parentheses =
0 gives critical numbers and
. The inequality is not true
at either of the critical numbers. We
can ignore the positive critical number
since x is negative for this case, and
so we take 0 as the upper critical
number. So the critical numbers are
and 0.
If x < , say x=-1, the
inequality becomes
[2(-1)-3][3(-1)+1] < 0
(-2-3)(-3+1) < 0
(-5)(-2) < 0
10 < 0
That's false so the inequality does not
hold when x is less than
If x is between the two critical
numbers, say x=-.01, the inequality
becomes
[2(-.01)-3][3(0)+1] < 0
(-3.02)(1) < 0
-3.02 < 0
That's true, so Case 3 is true
when (,0).
So for case 3, the inequality has
solution set
(,0)
Answer: The solution set for the given
inequality is:
(,0) U (,∞)
Edwin