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

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Owlcoffee Group TitleBest ResponseYou've already chosen the best response.0
my objectuve is to simply it.
 6 months ago

sweetburger Group TitleBest ResponseYou've already chosen the best response.0
get those denominators the same
 6 months ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Are you trying to solve the equation for x?
 6 months ago

Owlcoffee Group TitleBest ResponseYou've already chosen the best response.0
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.
 6 months ago

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

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

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