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

1. Owlcoffee

my objectuve is to simply it.

2. sweetburger

get those denominators the same

3. Owlcoffee

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

4. myininaya

Are you trying to solve the equation for x?

5. Owlcoffee

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. myininaya

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

7. Owlcoffee

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

8. Owlcoffee

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