Owlcoffee
  • Owlcoffee
so I was working on a math exercise and I got a little stuck on this part: \[x ^{2}+(\frac{ (2x- \lambda)^{2} }{ 16 })+\frac{ (- \lambda - 4)x }{ 2 }+\frac{ (- \lambda -16)(2x- \lambda) }{ 8 }+ \lambda = 0\]
Mathematics
jamiebookeater
  • jamiebookeater
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Owlcoffee
  • Owlcoffee
my objectuve is to simply it.
sweetburger
  • sweetburger
get those denominators the same
Owlcoffee
  • Owlcoffee
yes, I multiplied the whole equation by 16: \[16x ^{2}+(2x- \lambda)^{2}+(8)(x)(- \lambda - 4)+(2)(- \lambda -16)(2x- \lambda)+16 \lambda=0\]

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myininaya
  • myininaya
Are you trying to solve the equation for x?
Owlcoffee
  • Owlcoffee
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.
myininaya
  • myininaya
so you are trying to set it up as ax^2+bx+c=0?
Owlcoffee
  • Owlcoffee
yes. If possibly, I'd have to search another way of I can't set it up like that.
Owlcoffee
  • Owlcoffee
okay, I could pretty much do it: \[20x ^{2}+(-16 \lambda -96)x+( \lambda ^{2}+48 \lambda )=0\] :)

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