# Thread: Help with solving inequalities

1. ## Help with solving inequalities

1) $\displaystyle x^2 + 4x - 32 < 0$

I read in the book that you're supposed to factor and then plug in some "key" numbers that will cause the equation to equal zero (if I'm understanding this correctly). The only thing that I could come up with was:

(x+8) (x-4), but plugging in an '8' in the equation doesn't equal zero. How would I solve this? Thanks in advance.

2) This is an inequality that's just one variable (not sure how to proceed):

$\displaystyle \frac{x+1}{2x-3}$ ≥ 0

2. Here is the idea. If $\displaystyle x<-8$ then the factor $\displaystyle (x+8)$ is negative and for $\displaystyle x>-8$ that factor is positive.
If $\displaystyle x<4$ then the factor $\displaystyle (x-4)$ is negative and for $\displaystyle x>4$ that factor is positive.

Now if $\displaystyle -8<x<4$ we have one factor positive and the other negative, hence their product is negative. So have we found the complete solution?

3. Originally Posted by Plato
Here is the idea. If $\displaystyle x<-8$ then the factor $\displaystyle (x+8)$ is negative and for $\displaystyle x>-8$ that factor is positive.
If $\displaystyle x<4$ then the factor $\displaystyle (x-4)$ is negative and for $\displaystyle x>4$ that factor is positive.

Now if $\displaystyle -8<x<4$ we have one factor positive and the other negative, hence their product is negative. So have we found the complete solution?
Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...

4. Originally Posted by lilrhino
Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...
I need some guidance on this please...

5. Originally Posted by lilrhino
Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...
-8 is the point you should be looking at. Plato's solution is complete. Look at the graph.

6. Originally Posted by Jhevon
-8 is the point you should be looking at. Plato's solution is complete. Look at the graph.
Thanks Plato & Jhevon!

It helps to graph it first (I see). What approach should I use in addition to graphing to solve these sort of inequalities? Do I just use a test number from each?

Also, how should I proceed to solve my 2nd inequality?

7. Originally Posted by lilrhino
1) $\displaystyle x^2 + 4x - 32 < 0$
First, note that $\displaystyle x^2+4x-32<0\iff(x-4)(x+8)<0$

Now you have to analyze two possibilities:

$\displaystyle \blacktriangleright~(x-4)>0~~\wedge~~(x+8)<0$

$\displaystyle \blacktriangleright~(x-4)<0~~\wedge~~(x+8)>0$

After, find the intersection of the first one and do the same for the second one. Finally, the intervals that you'll have, join them and that's the answer.

Apply the same idea with the next exercise, but remember that you cannot take the denominator equal to 0.

8. Originally Posted by Krizalid
First, note that $\displaystyle x^2+4x-32<0\iff(x-4)(x+8)<0$

Now you have to analyze two possibilities:

$\displaystyle \blacktriangleright~(x-4)>0~~\wedge~~(x+8)<0$

$\displaystyle \blacktriangleright~(x-4)<0~~\wedge~~(x+8)>0$

After, find the intersection of the first one and do the same for the second one. Finally, the intervals that you'll have, join them and that's the answer.

Apply the same idea with the next exercise, but remember that you cannot take the denominator equal to 0.
I'll give it a try. Thanks for the explanation.

9. Originally Posted by lilrhino
Thanks Plato & Jhevon!

It helps to graph it first (I see). What approach should I use in addition to graphing to solve these sort of inequalities? Do I just use a test number from each?
i usually use a number-line approach. write the roots on the number line, in this case that would be -8 and +4, then test numbers in the different regions separated by the roots. so you would test like -9, 0 and 5 by plugging them into the original equation to see where the function would be negative or positive. you would realize that it is negative in the region between -8 and +4, which is the answer.

10. By the way lilrhino, use \ge to generate $\displaystyle a\ge0$, for example

11. Originally Posted by Krizalid
By the way lilrhino, use \ge to generate $\displaystyle a\ge0$, for example
i always use \geq, that works as well.

12. Originally Posted by Jhevon
i usually use a number-line approach. write the roots on the number line, in this case that would be -8 and +4, then test numbers in the different regions separated by the roots. so you would test like -9, 0 and 5 by plugging them into the original equation to see where the function would be negative or positive. you would realize that it is negative in the region between -8 and +4, which is the answer.
I'll try the number line approach because using a table is lengthy. Thanks for the explanation Jhevon.

13. Originally Posted by Jhevon
i always use \geq, that works as well.
Haha

If you prefer too, use \geqslant.

14. Originally Posted by Jhevon
i always use \geq, that works as well.
Originally Posted by Jhevon
haha, it wasn't a preference. i learnt LaTex from the tutorial here, in the tutorial \geq is what they have
I guess I'd better review the tutorial because I haven't a clue what you two are talking about

15. Originally Posted by lilrhino
I guess I'd better review the tutorial because I haven't a clue what you two are talking about
the tutorial is here

click on the pdf file in the first post

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