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Math Help - finding x when it appears in the equation more than once

  1. #1
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    finding x when it appears in the equation more than once

    Hi, this is the first time I have come across a question where x has appeared in an equation more than once so I am a bit stumped as how to find it.

    55.17 = [(2x)^2]/[(2-x)(2-x)]
    x = ?

    I can simply it to:
    55.17 = (4x^2) / [(2-x)^2]
    55.17^(1/2) = 4x / (2-x)
    but still can't find x

    any help would be great
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by Siddy View Post
    Hi, this is the first time I have come across a question where x has appeared in an equation more than once so I am a bit stumped as how to find it.

    55.17 = [(2x)^2]/[(2-x)(2-x)]
    x = ?

    I can simply it to:
    55.17 = (4x^2) / [(2-x)^2]
    55.17^(1/2) = 4x / (2-x)
    but still can't find x

    any help would be great
    denote \sqrt{55.17}=k.

    Then k= \frac{4x}{2-x} \Rightarrow k(2-x)=4x \Rightarrow 2k-2x=4x \Rightarrow 2k=4x+2x \Rightarrow 2k=6x \Rightarrow k/3=x
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  3. #3
    Member anonimnystefy's Avatar
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    Re: finding x when it appears in the equation more than once

    55.17^(1/2) = 4x / (2-x)
    the OP is wrong here,it should be 2x/(2-x)
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  4. #4
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    Re: finding x when it appears in the equation more than once

    Thanks so much abhishekkgp,
    can you tell me how you did this bit:

    k(2 - x) = 4x => 2k - 2x = 4x ?

    I thought it would be:

    k(2 - x) = 4x => 2k - kx = 4x
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  5. #5
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    Re: finding x when it appears in the equation more than once

    Thanks anonimnystefy, i forgot about square rooting the 2 there.
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  6. #6
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by Siddy View Post
    Thanks so much abhishekkgp,
    can you tell me how you did this bit:

    k(2 - x) = 4x => 2k - 2x = 4x ?

    I thought it would be:

    k(2 - x) = 4x => 2k - kx = 4x
    Yes, you are right. That must have been a typo.

    However, you are not done. from k= \frac{4x^2}{(2- x)^2} you can get both \sqrt{k}= \frac{2x}{2- x} and \sqrt{k}= \frac{2x}{x- 2}.
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  7. #7
    Senior Member abhishekkgp's Avatar
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by Siddy View Post
    Thanks so much abhishekkgp,
    can you tell me how you did this bit:

    k(2 - x) = 4x => 2k - 2x = 4x ?

    I thought it would be:

    k(2 - x) = 4x => 2k - kx = 4x
    my mistake!!the correct thing is k(2-x)=2k-kx.
    SORRY!!
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  8. #8
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    Re: finding x when it appears in the equation more than once

    Thanks HallsofIvy,

    solution 1.

    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    [K^1/2 (2 - x )]/2 = x
    (2.K^1/2 - x.K^1/2)/2 = x
    (2.K^1/2)/2 = x + x.K^1/2
    (2.K^1/2)/2.K^1/2 = 2x
    (2.K^1/2)/4.K^1/2 = x
    1/2 = x


    solution 2.
    K^1/2 = 2x / x - 2
    K^1/2 (x - 2) = 2x
    x.K^1/2 - 2K^1/2 = 2x
    -2K^1/2 = 2x - x.K^1/2
    (-2K^1/2) / K^1/2 = 2x - x
    -K^1/2 = x

    The answer can not be negative therefore solution 1 must be correct (as long as my working out is right).

    EDIT: actually i think i must have done something wrong because the answer should utilize K, but i cancelled it out. Because the answer needs to be different for different values of K.
    Can you guys point out my error?
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  9. #9
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by abhishekkgp View Post
    my mistake!!the correct thing is k(2-x)=2k-kx.
    SORRY!!
    No worries abhishekkgp, you still got me going in the right direction
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  10. #10
    Senior Member abhishekkgp's Avatar
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by Siddy View Post
    Thanks HallsofIvy,

    solution 1.

    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    [K^1/2 (2 - x )]/2 = x
    (2.K^1/2 - x.K^1/2)/2 = x
    (2.K^1/2)/2 = x + x.K^1/2 shouldn't this be x+(1/2)(x.sqrt(k))??
    (2.K^1/2)/2.K^1/2 = 2x
    (2.K^1/2)/4.K^1/2 = x
    1/2 = x


    solution 2.
    K^1/2 = 2x / x - 2
    K^1/2 (x - 2) = 2x
    x.K^1/2 - 2K^1/2 = 2x
    -2K^1/2 = 2x - x.K^1/2
    (-2K^1/2) / K^1/2 = 2x - x
    -K^1/2 = x

    The answer can not be negative therefore solution 1 must be correct (as long as my working out is right).

    EDIT: actually i think i must have done something wrong because the answer should utilize K, but i cancelled it out. Because the answer needs to be different for different values of K.
    Can you guys point out my error?
    ...
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  11. #11
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    Re: finding x when it appears in the equation more than once

    Thanks again abhishekkgp,

    solution 1.

    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    [K^1/2 (2 - x )]/2 = x
    (2.K^1/2 - x.K^1/2)/2 = x
    (2.K^1/2)/2 = x + (x.K^1/2)/2
    (4.K^1/2)/2 = 2x + x.K^1/2 <- but I still get stuck and can't separate the x's and the K's...


    I tried it again and went this way:
    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    K^1/2 (2 - x ) = 2x
    2.K^1/2 - x.K^1/2 = 2x
    2 - x = 2x / K^1/2
    2 = (2x + x) / K^1/2
    2.K^1/2 = 3x
    (2.K^1/2) / 3 = x

    Could that be right?

    Also, if it is right, how come I am stuck in the first part in this post, is there a trick I am missing?
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  12. #12
    Senior Member abhishekkgp's Avatar
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    Re: finding x when it appears in the equation more than once

    Quote Originally Posted by Siddy View Post
    Thanks again abhishekkgp,

    solution 1.

    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    [K^1/2 (2 - x )]/2 = x
    (2.K^1/2 - x.K^1/2)/2 = x
    (2.K^1/2)/2 = x + (x.K^1/2)/2
    (4.K^1/2)/2 = 2x + x.K^1/2 <- but I still get stuck and can't separate the x's and the K's...
    this gives 2.sqrt(k)=x(2+sqrt(k)) which gives x=(2.sqrt(k))/(2+sqrt(k))


    I tried it again and went this way:
    K^1/2 = 2x / 2 - x
    K^1/2 (2 - x) = 2x
    K^1/2 (2 - x ) = 2x
    2.K^1/2 - x.K^1/2 = 2x
    2 - x = 2x / K^1/2
    2 = (2x + x) / K^1/2
    2.K^1/2 = 3x
    (2.K^1/2) / 3 = x

    Could that be right?

    Also, if it is right, how come I am stuck in the first part in this post, is there a trick I am missing?
    ...
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