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

[norvald]

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 ?

 

[norvald]

Isolate a varaible in multi-variable function

Hi eveyone

I need some help isolate the variable l in this following expression

g and v are constants

Thank you very much

 

[Dr.Peterson]

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.

 

[JeffM]

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.