Question

Solve for x:4(x-1/x)^2-4(x+1/x)+1=0

Answer 1

let (x-1/x)=y
4y2-4y+1=0
4y2-2y-2y+1=0
2y(2y-1)-1(2y-1)=0
therefore,(2y-1)(2y-1)=0
hence y=1/2
putting value of y
x-1/x=1/2                                      [by cross multiplication]
2x-2=x
2x-x=2
x=2
Hence found.     :)
 

Answer 2

It is not clear if there is a typing mistake in the qn... whether it is supposed to be a simple one or tough one.

From the given equation it is clear that,  if  x=a is a solution, then x = 1/a is also a solution.

possibility 1:

$4 (x-\frac{1}{x})^2-4(x-\frac{1}{x}) + 1 =0\\\\Then\ x-\frac{1}{x}=\frac{4+-\sqrt{4^2-4*4}}{8}=\frac{1}{2}\\\\So\ 2x^2-x-2=0\\\\So: x=\frac{1+\sqrt{17}}{4},\ or, \frac{1-\sqrt{17}}{4}, ANSWER.$

Possibility 2:

$4 (x-\frac{1}{x})^2-4(x+\frac{1}{x}) + 1 =0,\ \ --(1)\\\\4(x^2+\frac{1}{x^2}-2)-4x-\frac{4}{x}+1=0\\\\4x^4-4x^3-7x^2-4x+4=0\ \ --(2)$

By checking for x = 0, 1, 2... we find that x = 2 satisfies the equation. So as written above, x = 1/2 is also a solution.  Let us write it as :

4 (x - 2) (x - 1/2) (x² + b x + 1) = 0
( 2x² - 5 x + 2) (2 x² + b x + 2) = 0    --(3)

Compare the coefficients of x³, x², and x in (2) and (3) to get
       b = 3

We find the roots of    2 x² + 3 x + 2 = 0. There  are no real roots as the discriminant is negative.

Answer :    x = 2 and 1/2.