Solve sqrt{3}/4 + sqrt{3}/(8g)*(-6g^2L+2gv^2)/(-3g^2L^2+2gv^2L+v^4)=0 for L

norvald

New member
Joined
Oct 24, 2018
Messages
2
Hi everyone

I need some help for this problem

I need to isolate l in this following expression but I find it hard

. . . . .\(\displaystyle \dfrac{\sqrt{3\,}}{4}\, +\, \dfrac{\sqrt{3\,}}{8g}\, \cdot\, \dfrac{(-6g^2 l\, +\, 2gv^2)}{-3g^2 l^2\, +\, 2gv^2 l\, +\, v^4}\, =\, 0\)

NB : g and v are constants
Can you help me ?
 

Attachments

  • JPEG1.jpg
    JPEG1.jpg
    24.2 KB · Views: 4
Last edited by a moderator:
Isolate a varaible in multi-variable function

Hi eveyone

I need some help isolate the variable l in this following expression
JPEG1.jpg
g and v are constants

Thank you very much
 
I need to isolate l in this following expression but I find it hard

. . . . .\(\displaystyle \dfrac{\sqrt{3\,}}{4}\, +\, \dfrac{\sqrt{3\,}}{8g}\, \cdot\, \dfrac{(-6g^2 l\, +\, 2gv^2)}{-3g^2 l^2\, +\, 2gv^2 l\, +\, v^4}\, =\, 0\)

g and v are constants
Start by dividing both sides by sqrt(3), which will make things look a lot simpler; then multiply both terms by the LCD, which is 8g times the trinomial. You will then have a quadratic equation, which can be solved by the quadratic formula.

If you need additional help, please show your work, so we can see where you are having trouble.
 
Last edited by a moderator:
I need to isolate l in this following expression but I find it hard

. . . . .\(\displaystyle \dfrac{\sqrt{3\,}}{4}\, +\, \dfrac{\sqrt{3\,}}{8g}\, \cdot\, \dfrac{(-6g^2 l\, +\, 2gv^2)}{-3g^2 l^2\, +\, 2gv^2 l\, +\, v^4}\, =\, 0\)

g and v are constants

Thank you very much
Dr. Peterson's first step is what I would do first as well except I would divide by

\(\displaystyle 2g\sqrt{3} \ \because \ g \ne 0.\)

\(\displaystyle \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{8g} * \dfrac{-\ 6g^2L + 2gv^2}{-\ 3g^2L^2 + 2gv^2L + v^4} = 0 \implies \)

\(\displaystyle \dfrac{1}{8g} + \dfrac{1}{16g^2} * \dfrac{-\ 3gL + 2v^2}{-\ 3g^2L^2 + 2gv^2L+ v^4} = 0.\)

Then I would isolate terms in L on one side of the equation and simplify before dealing with the quadratic.

\(\displaystyle \dfrac{1}{8g} + \dfrac{1}{16g^2} * \dfrac{-\ 3gL + 2v^2}{-\ 3g^2L^2 + 2gv^2L + v^4} = 0 \implies \\ \dfrac{-\ 3gL + 2v^2}{-\ 3g^2L^2 + 2gv^2L + v^4} = -\ \dfrac{16g^2}{8g} = -\ 2g.\)

Slightly different approach. I just like to simplify as much as possible before messing with the quadratic formula.
 
Last edited by a moderator:
Top