Math, asked by bdhxjkahagg, 2 months ago

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

Answers

Answered by VelvetRosee
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]

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