How to solve this? 4/(x^2-3x+2) - 3/(2x^2-6x+1) +1 = 0

How to solve this? 4/(x^2-3x+2) - 3/(2x^2-6x+1) +1 = 0

RESULT IS : X1=0 , x2=3 , x3,4=±i√11)/2

4/(x^2-3x+2) - 3/(2x^2-6x+1) +1 = 0

let 2x^2-6x+1 = Y, then

2*4/2*(x^2-3x+2) - 3/(2x^2-6x+1) +1 = 0

2*4/(2x^2-6x+4) - 3/(2x^2-6x+1) +1 = 0

8/(Y+3) - 3/Y +1 = 0

8Y/Y(Y+3) - 3(Y+3)/Y(Y+3) + Y*(Y+3)/Y*(Y+3) = 0

(8Y - 3Y - 9 + Y^2 + 3Y) / Y*(Y+3) = 0

(Y^2 + 8Y - 9) / Y*(Y+3) = 0

(Y-1)(Y+9) / Y*(Y+3) = 0

Y = 1 or -9

When Y = 1

2x^2-6x+1 = 1

2x^2-6x=0

x(2x - 6) = 0

x = 0 or 3


When Y = -9

2x^2-6x+1 = -9

2x^2-6x+10=0

x^2-3x+5=0 (divided by 2)

x = [-(-3) ± √((-3)^2 - 4*1 5)] /2

x = [3 ± √((9 - 20)] /2

x = [3 ± √(-11)] /2

x = (3 ± i√11) /2 <=== your answers of x3 and x4 don't seem to be correct!!

Multipy by the LCM which is ( x^2 - 3x + 2 ) * ( 2x^2 - 6x +1 )

4 ( 2x^2 - 6x + 1 ) - 3 ( x^2 - 3x + 2 ) +1 ( x^2 -3x +2 ) * ( 2x^2 - 6x + 1 ) = 0

Either you can factorise and find the solution from here .. or expand it by multiplying and solving then same