Question

Solve: x^4 – 2x^3 – x^2 – 2x + 1 = 0 and please don't scam.
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Answer 1

Answer:

$x=$ [(3±√5)/2] ;  $x=$ [(-1±√3i)/2]

Step-by-step explanation:

given equation :

$x^4 - 2x^3-x^2-2x+1=0$

let's take '$x^2$' common from above equation

⇒$x^2( x^2-2x-1-(2/x)+1/x^2 =0$

it is converted as :

⇒ $x^2-2x-1-(2/x)+1/x^2) = 0$

rearrange the above equation;

⇒ $[x^2+(1/x^2)]-2[x+(1/x)]-1 = 0$

let $[x+(1/x)]$ = A ;

squaring on both sides , we get ;

$x^2 +(1/x^2) +2$ = A²;

$x^2+1/x^2 = A^2 -2$

substitute value of A and A²  in above equation $[x^2+(1/x^2)]-2[x+(1/x)]-1 = 0$

A² - 2  - 2A - 1 =0 ; A² - 2A -3 =0

A² - 3A + A -3 =0;

(A-3)(A+1) =0

so A can be 3 or (-1)

so $[x+(1/x)]$  can be 3 or (-1)

let $[x+(1/x)]$ = 3;

$x^2-3x+1=0;$

$x=$ [(3±√5)/2]

if we consider $[x+(1/x)]$ = -1;

$x^2+x+1=0;$

$x=$ [(-1±√3i)/2]