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Owlcoffee Group Title

so I was working on a math exercise and I got a little stuck on this part: \[x ^{2}+(\frac{ (2x- \lambda)^{2} }{ 16 })+\frac{ (- \lambda - 4)x }{ 2 }+\frac{ (- \lambda -16)(2x- \lambda) }{ 8 }+ \lambda = 0\]

  • 5 months ago
  • 5 months ago

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  1. Owlcoffee Group Title
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    my objectuve is to simply it.

    • 5 months ago
  2. sweetburger Group Title
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    get those denominators the same

    • 5 months ago
  3. Owlcoffee Group Title
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    yes, I multiplied the whole equation by 16: \[16x ^{2}+(2x- \lambda)^{2}+(8)(x)(- \lambda - 4)+(2)(- \lambda -16)(2x- \lambda)+16 \lambda=0\]

    • 5 months ago
  4. myininaya Group Title
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    Are you trying to solve the equation for x?

    • 5 months ago
  5. Owlcoffee Group Title
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    No no, The original excercise is about some different. on the process I got that equation with two variables and what I'm trying to do is form a quadratic equation with it. in terms of x. Lambda is just a arbitrary constant.

    • 5 months ago
  6. myininaya Group Title
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    so you are trying to set it up as ax^2+bx+c=0?

    • 5 months ago
  7. Owlcoffee Group Title
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    yes. If possibly, I'd have to search another way of I can't set it up like that.

    • 5 months ago
  8. Owlcoffee Group Title
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    okay, I could pretty much do it: \[20x ^{2}+(-16 \lambda -96)x+( \lambda ^{2}+48 \lambda )=0\] :)

    • 5 months ago
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