Solve Inequalities algebraically

[Panphobia]

Homework Statement

The Attempt at a Solution

So when I started this question, I knew that 2r +3 > 0 in order for this whole thing to work so r > -3/2, and I know that if the middle term has to be more than 0, than only one more of the others is going to be negative, for the first term i found that r < 4/3 and the third was r < 2, so I know that the answer should be (4/3, 2), but how do I prove that algebraically?

 

[eumyang]

So when I started this question, I knew that 2r +3 > 0 in order for this whole thing to work so r > -3/2, and I know that if the middle term has to be more than 0, than only one more of the others is going to be negative, for the first term i found that r < 4/3 and the third was r < 2, so I know that the answer should be (4/3, 2), but how do I prove that algebraically?

I wouldn't solve it "algebraically." I would solve it by making a two-column table. First, find values of r that would make the left side 0 or undefined (I call these critical values), which appears you have already done. Then in the first column list all critical values and all intervals bounded by these critical values.

Test interval|Value
-------------+----------
(-inf, -3/2) |undefined
r = -3/2     |0
(-3/2, 4/3)  |
r = 4/3      |0
(4/3, 2)     |
r = 2        |0
(2, inf)     |

In the second column, you fill in 0, undefined, positive, or negative. For the remaining intervals, pick a test value to plug into the inequality and you will eventually determine the solution set.

 

[Panphobia]

ohhhh ok, that makes sense. thank you.

 

[HallsofIvy]

Since the square root, $\sqrt{2r+ 3}$, is, by definition, nonnegative, you can (almost) ignore it. The product of the two numbers, 3r- 4 and $\sqrt[3]{r- 2}$, will be negative if and only if the two factors are of opposite order. In other words, either

1) 3r- 4> 0 and $\sqrt[3]{r- 2}< 0$ or
2) 3r- 4< 0 and $\sqrt[3]{r- 2}> 0$

Similarly, a positive number, cubed, is positive and a negative number, cubed, is negative. So that can be reduced to
1) 3r- 4> 0 and r- 2< 0 or
2) 3r- 4< 0 and r- 2> 0.

3r- 4> 0 for r> 4/3 and r- 2< 0 for r< 3. Those will both be true for 4/3< r< 3.
$\sqrt{2r+ 3}= 0$ only for r= -3/2 and that does not lie between 4/3 and 3.

(1) is true for 4/3< r< 3.

3r- r< 0 for r< 4/3 and r- 2> 0 for r> 3. (2) is never true.