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

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Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0my objectuve is to simply it.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.0get those denominators the same

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0yes, I multiplied the whole equation by 16: \[16x ^{2}+(2x \lambda)^{2}+(8)(x)( \lambda  4)+(2)( \lambda 16)(2x \lambda)+16 \lambda=0\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Are you trying to solve the equation for x?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0No 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
 one year ago
Best ResponseYou've already chosen the best response.2so you are trying to set it up as ax^2+bx+c=0?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0yes. If possibly, I'd have to search another way of I can't set it up like that.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0okay, I could pretty much do it: \[20x ^{2}+(16 \lambda 96)x+( \lambda ^{2}+48 \lambda )=0\] :)
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