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

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

Owlcoffee
 9 months 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
 9 months ago
Best ResponseYou've already chosen the best response.2Are you trying to solve the equation for x?

Owlcoffee
 9 months 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
 9 months ago
Best ResponseYou've already chosen the best response.2so you are trying to set it up as ax^2+bx+c=0?

Owlcoffee
 9 months 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
 9 months 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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