# Thread: Substition transformation to quadratic form

1. ## Substition transformation to quadratic form

$
\frac{\1}{2x-1}^2 + \frac{\1}{2x-1}^2 -12=0
$

Apologies for the sloppy LaTex formatting. There should be parentheses around the fractions indicating that both need to be squared.

I think my first issue lies with squaring the fraction. If I could do that I'd be on the way. I don't normally ask to be spoon fed an answer but in this case... Thanks in advance for any all help.

2. ## Re: Substition transformation to quadratic form

$\displaystyle \left(\frac{1}{2x-1}\right)^2+\left(\frac{1}{2x-1}\right)^2-12=0$

$\displaystyle 2\left(\frac{1}{2x-1}\right)^2-12=0$

$\displaystyle 2\left(\frac{1}{2x-1}\right)^2=12$

Now divide both sides by 2 and take the square root, that should get you pretty close to the end.

What do you get?

3. ## Re: Substition transformation to quadratic form

Since you titled this "Substitution transformation to quadratic form", use the substitution $u= \frac{1}{2x- 1}$ so your equation becomes $u^2+ u^2- 12= 2u^2- 12= 0$