# Thread: Inequalities - due tomorrow!

1. ## Inequalities - due tomorrow!

Hey, this assignment is due tomorrow and we never focused on inequalities before, so I need a hand with these couple problems. Also, I don't really quite know how to use most of the symbols, so please bear with me.

These have to be solved and the answer stated in interval notation.

a. 45x^4 - 20x^2 greater than 0

b. -2/(x+5) greater than 0

c. (x^2 + 8x + 15)/(x^2 - 1) greater than or equal to 0

d. X^3 - 63 less than or equal to 1

I know a little bit about key numbers, number lines, etc. The main purpose for this is to confirm the answers i have already, especially since she doesn't give partial credit!

2. Hello, leviathanwave!

Here are some of them . . .

$\displaystyle a)\;\;45x^4 - 20x^2 \:> \:0$

Factor: .$\displaystyle 5x^2(9x^2 - 4) \:> \:0$

For $\displaystyle x \neq 0,\;5x^2$ is always positive.
Hence: $\displaystyle 9x^2 - 4$ must be positive.

We have: .$\displaystyle 9x^2 - 4 \:>\:0\quad\Rightarrow\quad x^2 \:> \:\frac{4}{9}\quad\Rightarrow\quad|x| > \frac{2}{3}$

Solution: .$\displaystyle \left(\text{-}\infty,\,\text{-}\frac{2}{3}\right) \:\cup \:\left(\frac{2}{3},\,\infty\right)$

$\displaystyle b)\;\;\frac{-2}{x+5} \:> \:0$

For the fraction to be positive, its denominator must also be negative.

We have: .$\displaystyle x + 5 \:< \:0\quad\Rightarrow\quad x \:< \:-5$

Solution: .$\displaystyle (\text{-}\infty,\,\text{-}5)$

$\displaystyle c)\;\;\frac{x^2 + 8x + 15}{x^2 - 1} \:\geq \:0$

This is a messy one . . . I'll let someone else explain it.

$\displaystyle d)\;\;x^3 - 63\:\leq \:1$

We have: .$\displaystyle x^3 \:\leq \:64\quad\Rightarrow\quad x \:\leq \:4$

Solution: .$\displaystyle (-\infty,\,4]$

3. For c: determine where the numerator is positive, zero and negative and do the same for the denominator.
The inequality then holds wherever the numerator (only) is 0 or where numerator and denominator have the same sign.

4. Originally Posted by leviathanwave
c. (x^2 + 8x + 15)/(x^2 - 1) greater than or equal to 0
First we exclude $\displaystyle x=\pm 1$.

There are 2 possibilities,
$\displaystyle x^2-1<0$
$\displaystyle x^2-1>0$

Let us first assume the first case,
$\displaystyle x^2-1<0$
$\displaystyle x^2<1$
$\displaystyle |x|<1$ *IMPORTANT*
Multiplying both sides by the denominator will flip the inequality sign because it is negative,
$\displaystyle x^2+8x+15\leq 0$
$\displaystyle x^2+8x \leq -15$
$\displaystyle x^2+8x+16 \leq -15+16$
$\displaystyle (x+4)^2 \leq 1$
$\displaystyle -1\leq x+4 \leq 1$
$\displaystyle 3\leq x\leq 5$
Look at the important message, we assumed that $\displaystyle |x|<1$ the only ones that work are $\displaystyle -1<x<1$. Thus, that is one solution,
$\displaystyle x=(-1,1)$

Let us assume the second case,
$\displaystyle x^2-1>0$
$\displaystyle x^2>1$
$\displaystyle x>1 \mbox{ or }x<-1$ *IMPORTANT*
Multiplying both sides by the denominator will not change the inequality sign for it is positive,
$\displaystyle x^2+8x+15\geq 0$
$\displaystyle x^2+8x+16\geq 1$
$\displaystyle (x+4)^2\geq 1$
$\displaystyle x+4 \geq 1 \mbox{ or }x+4\leq -1$
$\displaystyle x \geq -3 \mbox{ or }x\leq -5$
It needs to agree with the important interval.
Thus, $\displaystyle -3\leq x< -1 \mbox { or } x>1 \mbox{ or } x\leq -5$
The solution is hence,
$\displaystyle x=[-3,-1)\cup (1,\infty) \cup (-\infty, -5]$

Thus, the full solution (from first problem) is,
$\displaystyle x=[-3,-1)\cup (1,\infty) \cup (-\infty, -5] \cup (-1,1)$

5. Originally Posted by ThePerfectHacker
Let us first assume the first case,
$\displaystyle x^2-1<0$
$\displaystyle x^2<1$
$\displaystyle |x|<1$ *IMPORTANT*
Multiplying both sides by the denominator will flip the inequality sign because it is negative,
$\displaystyle x^2+8x+15\leq 0$
$\displaystyle x^2+8x \leq -15$
$\displaystyle x^2+8x+16 \leq -15+16$
$\displaystyle (x+4)^2 \leq 1$
$\displaystyle -1\leq x+4 \leq 1$
$\displaystyle 3\leq x\leq 5$
Look at the important message, we assumed that $\displaystyle |x|<1$ the only ones that work are $\displaystyle -1<x<1$. Thus, that is one solution,
$\displaystyle x=(-1,1)$
If you find that x has to be in [3,5] when you're in the case |x| < 1, then there are no solutions.
So in your final solution, the interval (-1,1) shouldn't be there. The rest is correct.

6. Hey guys, thanks so much for the help. Yeah, b and c were the ones i didnt actually have the right answers for. Thanks again for the help, you saved my life!