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Math Help - Illegal Values

  1. #1
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    Illegal Values

    I must be growing senile in my old age. Else how can you explain I have simply forgotten how to do this problem

    Find the illegal values of c in the multiplication statement c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
    a.) c=7, c=3, c=-2, c=-5 b.) c=-7, c=-3, c=-2, and c=5 c.) c=-7, c=-3, c=2, and c=5 d.) c=-7 and c=-3

    I reached as far as this ((c+2)(c+1)/(c+7)(c+3))

    Now what? Something has to equal 0 except I dont know if its the numerator or the denominator.

    Similarly
    Find the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8.
    a.) b=-5 and 2 b.) b=-2 and -4 c.) b=-2 and 4 d.) b=-5, -2, 2 and 4

    Help Please
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  2. #2
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    Quote Originally Posted by askmemath
    I must be growing senile in my old age. Else how can you explain I have simply forgotten how to do this problem

    Find the illegal values of c in the multiplication statement c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
    a.) c=7, c=3, c=-2, c=-5 b.) c=-7, c=-3, c=-2, and c=5 c.) c=-7, c=-3, c=2, and c=5 d.) c=-7 and c=-3

    I reached as far as this ((c+2)(c+1)/(c+7)(c+3))

    Now what? Something has to equal 0 except I dont know if its the numerator or the denominator.

    Similarly
    Find the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8.
    a.) b=-5 and 2 b.) b=-2 and -4 c.) b=-2 and 4 d.) b=-5, -2, 2 and 4

    Help Please
    One of the rules of this forum is not to discuss something which is illegal


    Illegal values over here is where the function is undefined.
    Thus,
    \frac{c^2-3c-10}{c^2+5c-14} factors as,
    \frac{(c-5)(c+2)}{(c+7)(c-2)}.
    Notice that this is a fraction, and thus it is undefined where the denominator is zero. Thus, in this case,
    (c+7)(c-2)=0 which gives the solutions,
    c=-7,2.
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  3. #3
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    Thank you.

    I would like to ask whether it make a difference that the above term is also being multiplied by another expression since the question quite clearly says
    c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
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  4. #4
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    Quote Originally Posted by askmemath
    Thank you.

    I would like to ask whether it make a difference that the above term is also being multiplied by another expression since the question quite clearly says
    c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
    It does not matter, you always look at the bottom.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    It does not matter, you always look at the bottom.
    He is trying to tell you the question asked for the illegal values of c in:

    <br />
\frac{c^2-3c-10}{c^2+5c-14} \times \frac{c^2-c-2} {c^2-2c-15}<br />

    RonL
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  6. #6
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    So basically the final answer should be that the illegal values of C are -7 and -3
    Right?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by askmemath
    So basically the final answer should be that the illegal values of C are -7 and -3
    Right?
    Yes, if you are allowed to cancel the factors of the terms at the top and
    bottom which are equal, but strictly the illegal values are:

    c.) c=-7, c=-3, c=2, and c=5

    The ratio is indeterminate for c=2 and c=5, but the limits exist at these points.

    RonL
    Last edited by CaptainBlack; March 2nd 2006 at 09:29 AM.
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