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Thread: Question about comparison

  1. #1
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    Question about comparison

    Given is the function: f(x)= (x^2-4x) / (x^2+2)


    Asked:

    For which p ( element R ) contains the set of solutions of the equation f (x) = p exactly one element ?
    To give the found p the set of solutions.



    How do I solve this task correctly? Look, I can fill in for p, p=-1 , p=1 and p=2 .

    So I find the answers= p=-1 with x=1
    p=1 with with x=1/2
    p=2 with x=-2


    But is this the right way? I filled these values for p , because I saw this in the answer book.
    If I did not have the answer book, how do I know which values of p should I choose?
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  2. #2
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    Re: Question about comparison

    $\dfrac{x^2-4x}{x^2+2}-p = 0$

    $\dfrac{x^2-4x}{x^2+2} - \dfrac{p(x^2+2)}{x^2+2} = 0$

    $\dfrac{x^2(1-p)-4x-2p}{x^2+2}=0$

    Rational expression equals zero when the numerator equals zero ...

    $x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$

    to get a single solution for $x$, the discriminant must equal zero. Can you finish?
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    Re: Question about comparison

    Equivalently:
    \frac{x^2- 4x}{x^2+ 2}= p
    x^2- 4x= px^2+ 2p
    (1- p)x^2- 4x= 2p
    x^2- \frac{4}{1- p}x= \frac{2p}{1- p}

    Complete the square:
    x^2- \frac{4}{1- p)x+ \frac{4}{(1- p)^2= \frac{2p}{1- p}+ \frac{4}{(1- p)^2= \frac{2p- 2p^2+ 4}{(1- p)^2}
    (x- \frac{2}{1- p})^2= -2\frac{p^2- p- 2}{(1- p)^2}

    Solve by taking the square root of both sides. If the right side is negative, there is no root. If the right side is positive we can take both "+" and "-" and there are two real roots. If the right side is 0, there is only one root.
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    Re: Question about comparison

    Quote Originally Posted by skeeter View Post
    $\dfrac{x^2-4x}{x^2+2}-p = 0$

    $\dfrac{x^2-4x}{x^2+2} - \dfrac{p(x^2+2)}{x^2+2} = 0$

    $\dfrac{x^2(1-p)-4x-2p}{x^2+2}=0$

    Rational expression equals zero when the numerator equals zero ...

    $x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$

    to get a single solution for $x$, the discriminant must equal zero. Can you finish?


    If I calculate for the discriminant of this, I find: p=-1 and p =2 , I find two solutions, so the discriminant is not equal to zero?
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    Re: Question about comparison

    Quote Originally Posted by HallsofIvy View Post
    Equivalently:
    \frac{x^2- 4x}{x^2+ 2}= p
    x^2- 4x= px^2+ 2p
    (1- p)x^2- 4x= 2p
    x^2- \frac{4}{1- p}x= \frac{2p}{1- p}

    Complete the square:
    x^2- \frac{4}{1- p)x+ \frac{4}{(1- p)^2= \frac{2p}{1- p}+ \frac{4}{(1- p)^2= \frac{2p- 2p^2+ 4}{(1- p)^2}
    (x- \frac{2}{1- p})^2= -2\frac{p^2- p- 2}{(1- p)^2}

    Solve by taking the square root of both sides. If the right side is negative, there is no root. If the right side is positive we can take both "+" and "-" and there are two real roots. If the right side is 0, there is only one root.


    On the left I find: x=0 .

    On the right I find : p=-1 and p =2 . Is this what you meant ?
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    Re: Question about comparison

    Quote Originally Posted by westerwolde View Post
    Given is the function: f(x)= (x^2-4x) / (x^2+2)
    For which p ( element R ) contains the set of solutions of the equation f (x) = p exactly one element ?
    Quote Originally Posted by skeeter View Post
    $\dfrac{x^2-4x}{x^2+2}-p = 0$
    Rational expression equals zero when the numerator equals zero ...
    $x=\dfrac{4 \pm \sqrt{(-4)^2-4(1-p)(-2p)}}{2(1-p)}$
    to get a single solution for $x$, the discriminant must equal zero.
    Quote Originally Posted by westerwolde View Post
    On the left I find: x=0 .
    On the right I find : p=-1 and p =2 . Is this what you meant ?
    Yes those are the correct answers. Each of those values for $\bf{p}$ gives a single solution for the original question.
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  7. #7
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    Re: Question about comparison

    But I don't know what you mean by "x= 0". Those values of p make the part after \pm equal to 0 so that the single solution to the equation is x= \frac{4\pm 0}{2(1- p)}. If p= -1 that is x= \frac{4}{4}= 1 and if p= 2 that is x= \frac{4}{-2}= -2.
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    Re: Question about comparison

    Okay, this is clear. But I still do not understand how they come in p=1 with with x=1/2 in the answer book.. Or will this be a mistake in the answer book?
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  9. #9
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    Re: Question about comparison

    Quote Originally Posted by westerwolde View Post
    Okay, this is clear. But I still do not understand how they come in p=1 with with x=1/2 in the answer book.. Or will this be a mistake in the answer book?
    Finding the number of solutions of the equation f(x)=p is equivalent to finding the number of solutions of the equation

    x^2(1-p)-4x-2p=0

    Case 1. p=1 this is a linear equation which has a unique solution

    Case 2. p\neq 1 we have a quadratic equation whose discriminant must be 0 if we want to have exactly one solution
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