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Math Help - Polynomial inequalities

  1. #1
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    Polynomial inequalities

    I have a question I worked out the problems on the bottom but it says my answers are wrong. Can you please tell me what I did wrong?

    The first problem is the first attached file for it I got the answer [2,4] by doing the sign chart

    The second problem is the second file attached for it I got the answer [-4,0]U[8,13]

    The third problem is the third file attached and for it I got (-infinity,-6)U(6,infinity)

    The fourth problem is the fourth file attached and for that I got (-infinity,-13]U[4,infinity)

    All those answers are wrong, so can you tell me what I did wrong?
    Attached Thumbnails Attached Thumbnails Polynomial inequalities-85e224d291c1a2449b6d1855941cfe1.png   Polynomial inequalities-33f7698398bca17db6b95ab4913cbd1.png   Polynomial inequalities-b506b040173a14eb82fac74d091b121.png   Polynomial inequalities-91962860d9aaea51a6b786640234cc1.png  
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  2. #2
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    Quote Originally Posted by Girlaaaaaaaa View Post
    I have a question I worked out the problems on the bottom but it says my answers are wrong. Can you please tell me what I did wrong?

    The first problem is the first attached file for it I got the answer [2,4] by doing the sign chart

    The second problem is the second file attached for it I got the answer [-4,0]U[8,13]

    The third problem is the third file attached and for it I got (-infinity,-6)U(6,infinity)

    The fourth problem is the fourth file attached and for that I got (-infinity,-13]U[4,infinity)

    All those answers are wrong, so can you tell me what I did wrong?
    Most of your answers are wrong because you include the case that the denominator equals zero. For instance:

    The solution of the 1rst problem is (2, 4). With your solution you actually divide by zero

    I've got a total different solution with the 3rd problem: \frac74 < x < 6 or as an interval: \left(\frac74 , 6 \right)
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  3. #3
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    Quote Originally Posted by earboth View Post
    Most of your answers are wrong because you include the case that the denominator equals zero. For instance:

    The solution of the 1rst problem is (2, 4). With your solution you actually divide by zero

    I've got a total different solution with the 3rd problem: \frac74 < x < 6 or as an interval: \left(\frac74 , 6 \right)
    I just tried your answer for the third problem and it showed that it is incorrect also. Btw how did you get that answer?
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  4. #4
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    Quote Originally Posted by Girlaaaaaaaa View Post
    I just tried your answer for the third problem and it showed that it is incorrect also. Btw how did you get that answer?
    You are right - but I am right too :

    \frac1{x-6}>4 . You can multiply the inequaltity by (x-6) if you pay attention to the sign of the denominator:

    \frac1{x-6}>4~\iff~1>4x-24~\wedge~x-6>0~\iff~ \frac{25}4 > x ~\wedge x>6 that means \left(6~,~\frac{25}4\right)

    \frac1{x-6}>4~\iff~1<4x-24~\wedge~x-6<0~\iff~ \frac{25}4 < x ~\wedge x<6 that means there isn't any real x which satisfies this case.
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  5. #5
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    Hello,

    For the first one, you can say that the inverse function decreases. This means that for every x & y, if 1/x < 1/y, 1/(1/x)=x > 1/(1/y)=y

    So it's the same as solving x-4 > x-2, which is always false. So there is no solution...

    For the third one, it's the same method. If i'm not mistaking, the answer is x<25/4, excluding x=6 because it's a zero of the denominator.
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    Here a comment about the usual way this taught in North America.
    Solving \frac{1}{{x - 6}} > 4 students are encouraged to ‘compare to zero’.
    \frac{1}{{x - 6}} - 4 > 0 \Rightarrow \quad \frac{{25 - 4x}}{{x - 6}} > 0 \Rightarrow \quad \frac{{4x - 25}}{{x - 6}} < 0.
    It is easy to check to see what values that make the statement true.
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