Hi I'm kinda confused by these equations.
1.) 3[ (x+3) / (2x-1) ]^2 - 4(x+3) / (2x-1) + 1 = 0
2.) 36x^4/3 - 25x^2/3 + 4 = 0
Thanks!
3$\displaystyle (x+3 / 2x-1)^2 - 4(x+3 / 2x-1) + 1$
= $\displaystyle 3a^2 - 4a + 1 $ { replacing x$\displaystyle +3/2x-1) by a $}
= $\displaystyle 3a^2 - 3a - a + 1$
=
$\displaystyle 3a(a - 1) - 1(a-1)$= $\displaystyle (3a-1)(a-1)$
= $\displaystyle {3(x+3 / 2x-1) - 1}{3(x+3 / 2x-1) - 1} $
$\displaystyle
36x^4/3 - 25x^2/3 + 4 = 0
$
Or $\displaystyle 36a^2 - 25a + 4 = 0 $ {subtituting $\displaystyle x^2/3$ by a}
Or $\displaystyle 36a^2 - 16a -9a + 4=0$
Or $\displaystyle 4a(9a-4) - 1(9a-4)=0$
Or (4a-1)(9a-4)=0
Or $\displaystyle a=1/4, 4/9$
Or $\displaystyle x^2/3 = 1/4 , 4/9$
Or $\displaystyle x^1/3=1/2 , 2/3$
Or $\displaystyle x = 1/8 ,8/27. $