1. ## Take Home Quiz Questions Please Check

Wondering if i Did it right

1)
(x-3)/(x+4)≤0
x+4≤0
x≤-4
(-infin,-4]

2)
4x^2+2x+3=0
x=-2±√(4-48)/8
x=-2±√(-44)/8
x=-2±√(11*4)i/8
x=-2±2√(11)i/8
x=-1/4±√(11)i/4

2. for the first problem, x+4 cannot equal zero. So you have $\displaystyle (x-3)\le 0 \text{ or } x+4<0$

2.) $\displaystyle \sqrt{-44} = i\sqrt{44}$ - just realised you already did that...was having trouble reading the expression, and it's correct. You can however simplify further, by putting it as a single fraction over four.

3. hey thanks for the help maybe a dumb question but how come in the rational inequality the ≤ changes to > for the x+4?

4. Your welcome. The expression $\displaystyle {x+4}\le 0$ is false because the denominator of $\displaystyle \frac{x-3}{x+4}$ (or any fraction) cannot = 0...since division by zero doesn't make sense/is not defined.

5. Originally Posted by civiliam
Wondering if i Did it right

1)
(x-3)/(x+4)≤0
x+4≤0
x≤-4
(-infin,-4]

...
Here is another attempt:

$\displaystyle \dfrac{x-3}{x+4}\leq0~,~x \neq -4$ . The LHS is a negative quotient. A quotient is negative if numerator and denominator have different signs. That means you have:

$\displaystyle x-3\leq 0\ \wedge \ x+4>0~\vee~x-3 \geq 0\ \wedge \ x+4< 0$

$\displaystyle x\leq 3\ \wedge \ x>-4~\vee~x \geq 3\ \wedge \ x< -4$

$\displaystyle \boxed{-4<x\leq3}~\vee~ x \notin \mathbb{R}$

6. ah that makes sense thanks very much

7. Originally Posted by earboth
Here is another attempt:

$\displaystyle \dfrac{x-3}{x+4}\leq0~,~x \neq -4$ . The LHS is a negative quotient. A quotient is negative if numerator and denominator have different signs. That means you have:

$\displaystyle x-3\leq 0\ \wedge \ x+4>0~\vee~x-3 \geq 0\ \wedge \ x+4< 0$

$\displaystyle x\leq 3\ \wedge \ x>-4~\vee~x \geq 3\ \wedge \ x< -4$

$\displaystyle \boxed{-4<x\leq3}~\vee~ x \notin \mathbb{R}$
I did something similar, but got:

$\displaystyle x\in(-4,3]~\vee~x\in(-\infty , -4)\cup[-3,\infty)$

isn't $\displaystyle (-\infty , -4)\cup[-3,\infty)\subset\mathbb{R}$ ?

8. Originally Posted by Greengoblin
I did something similar, but got:

$\displaystyle x\in(-4,3]~\vee~x\in(-\infty , -4)\cup[-3,\infty)$

...
In my opinion the quoted line should read:

$\displaystyle x\in(-4,3]~\vee~x\in\underbrace{(-\infty , -4){\color{red}\cap}[{\color{red}\bold{+}}3,\infty)}_{empty}$

9. Sorry, yeh that should be plus three, and the intersection of the two is empty as you say, but I suppose you can look at it as not being defined for real variables at all, since for any x belongng to the intersection, the function isn't defined, or you can look at it that even though some real variables don't give a valid output, there is a domain (a subset) of the real numbers (the union) for which the function is defined. Well my opinion is that both points are as valid as the other, but maybe some situations might be better stated with one definition over the other.