so I was working on a math exercise and I got a little stuck on this part:
$$x ^{2}+(\frac{ (2x- \lambda)^{2} }{ 16 })+\frac{ (- \lambda - 4)x }{ 2 }+\frac{ (- \lambda -16)(2x- \lambda) }{ 8 }+ \lambda = 0$$

Question

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 · Answer

my objectuve is to simply it.

sweetburger · Answer

get those denominators the same

owlcoffee · Answer

yes, I multiplied the whole equation by 16:
$$16x ^{2}+(2x- \lambda)^{2}+(8)(x)(- \lambda - 4)+(2)(- \lambda -16)(2x- \lambda)+16 \lambda=0$$

myininaya · Answer

Are you trying to solve the equation for x?

owlcoffee · Answer

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 · Answer

so you are trying to set it up as ax^2+bx+c=0?

owlcoffee · Answer

yes. If possibly, I'd have to search another way of I can't set it up like that.

owlcoffee · Answer

okay, I could pretty much do it:
$$20x ^{2}+(-16 \lambda -96)x+( \lambda ^{2}+48 \lambda )=0$$
:)