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Math Help - Parametric system of equation problem

  1. #1
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    Parametric system of equation problem

    x^2+y^2=2a+4
    x+y=a+3

    find the meaning for "a" when system of equation has only 1 root

    please someone help. thanks

    (sorry for bad english)
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  2. #2
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    Re: Parametric system of equation problem

    Quote Originally Posted by Telo View Post
    x^2+y^2=2a+4
    x+y=a+3
    find the meaning for "a" when system of equation has only 1 root
    Solve x^2+(a+3-x)^2=2a+4 for x.
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  3. #3
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    Re: Parametric system of equation problem

    thanks for answer but i didnt get it, i have this: 2x^2-2ax+a^2+9=2a+4 what do i do next? how do i find out what meaning must "a" have? for 1 root of x
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  4. #4
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    Re: Parametric system of equation problem

    Do you not know how to solve a quadratic equation?

    But more to the point, that equation can be written 2x^2- 2ax+ (a^2-2a+ 5)= 0 Now, an equation like that will have only one root provided it is a double root. That is, if it is of the form 2(x- x_0)^2= 0.
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  5. #5
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    Re: Parametric system of equation problem

    Expanding on HallsofIvy's post, you have a double root if and only if the discriminant is zero, i.e.

    (-2a)^2 - 4(2)(a^2 - 2a + 5) = 0

    Solve the quadratic.
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  6. #6
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    Re: Parametric system of equation problem

    thank you very much! I've got a=-1
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  7. #7
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    Re: Parametric system of equation problem

    Hmm I'm not sure if -1 is correct...according to the quadratic I posted, it has two non-real roots for a.
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    Re: Parametric system of equation problem

    Quote Originally Posted by richard1234 View Post
    Hmm I'm not sure if -1 is correct...according to the quadratic I posted, it has two non-real roots for a.
    -1 should be correct. I get a single solution for the resulting system.

    My work disagrees slightly with HallsofIvy, however:

    \begin{align*}x^2+y^2&=2a+4\\x+y&=a+3\end{align*}

    \Rightarrow x^2 + \left(a-x+3\right)^2=2a+4

    \Rightarrow2x^2-2ax-6x+a^2+6a+9=2a+4

    \Rightarrow2x^2-2(a+3)x+\left(a^2+4a+5\right)=0

    For this quadratic to have a single root, the discriminant b^2-4ac must be 0:

    4(a+3)^2-8\left(a^2+4a+5\right)=0

    \Rightarrow a^2+2a+1=0

    \Rightarrow a=-1.

    Substituting this back into the original equations gives the system

    \begin{align*}x^2+y^2&=2\\x+y&=2\end{align*}

    So

    x^2 + (2-x)^2=2

    \Rightarrow x^2-2x+1=0

    \Rightarrow(x-1)^2=0

    \Rightarrow x=1

    \Rightarrow y=1

    A single solution, as desired.
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