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Math Help - Take Home Quiz Questions Please Check

  1. #1
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    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=-22√(11)i/8
    x=-1/4√(11)i/4
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  2. #2
    Member Greengoblin's Avatar
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    for the first problem, x+4 cannot equal zero. So you have (x-3)\le 0 \text{ or } x+4<0

    2.) \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.
    Last edited by Greengoblin; October 9th 2008 at 11:18 AM.
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  3. #3
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    hey thanks for the help maybe a dumb question but how come in the rational inequality the ≤ changes to > for the x+4?
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  4. #4
    Member Greengoblin's Avatar
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    Your welcome. The expression {x+4}\le 0 is false because the denominator of \frac{x-3}{x+4} (or any fraction) cannot = 0...since division by zero doesn't make sense/is not defined.
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  5. #5
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    Quote Originally Posted by civiliam View Post
    Wondering if i Did it right

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

    ...
    Here is another attempt:

    \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:

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

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

    \boxed{-4<x\leq3}~\vee~ x \notin \mathbb{R}
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  6. #6
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    ah that makes sense thanks very much
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  7. #7
    Member Greengoblin's Avatar
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    Quote Originally Posted by earboth View Post
    Here is another attempt:

    \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:

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

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

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

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

    isn't (-\infty , -4)\cup[-3,\infty)\subset\mathbb{R} ?
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  8. #8
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    Quote Originally Posted by Greengoblin View Post
    I did something similar, but got:

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

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

    x\in(-4,3]~\vee~x\in\underbrace{(-\infty , -4){\color{red}\cap}[{\color{red}\bold{+}}3,\infty)}_{empty}
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  9. #9
    Member Greengoblin's Avatar
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    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.
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