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

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

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

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

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

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