Quartic Equation

Algebra Level 3

x 4 − 4 x 3 + 5 x 2 − 4 x + 1 = 0 \large x^4 - 4x^3 + 5x^2 - 4x + 1 = 0

Find all the real values of x x satisfying the equation above.

0 0 3 ± 5 2 \frac{3\pm \sqrt5}2 6 + 10 2 \frac{6+\sqrt{10}}2 6 + 5 2 6 + \frac{\sqrt5}2

William Isoroku by William Isoroku , 25, USA

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2 solutions

Abdul Lah
Sep 29, 2014

divide each term by x^2 so we get x^2 - 4x + 5 - 4/x +1/x^2 = 0 now (x^2 +1/x^2) - 4( x +1/x) +5 = 0 now let x +1/x = y implies that x^2 +1/x^2 = y^2 - 2 now put in equation we get y^2 - 4 y +3 = 0 by solving we get y =3 , y=1 back substitute in x+1/x = y we have x^2-3x+1 = 0 using quadratic formula x = 3 plus minus sqrt 5 /2

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Tom Engelsman
May 15, 2016

Let the above polynomial be re-written as:

x 4 − 4 x 3 + 5 x 2 − 4 x + 1 = ( x 4 − 4 x 3 + 6 x 2 − 4 x + 1 ) − x 2 = ( x − 1 ) 4 − x 2 = 0 ; x^4 - 4x^3 + 5x^2 - 4x + 1 = (x^4 - 4x^3 + 6x^2 - 4x + 1) - x^2 = (x-1)^4 - x^2 = 0;

or ( x − 1 ) 4 = x 2 (x-1)^4 = x^2

or ( x − 1 ) 2 = ± x . (x-1)^2 = \pm x.

This now gives two separate quadratic equations:

x 2 − 3 x + 1 = 0 with roots x = [ 3 ± 5 ] / 2 ; x^2 - 3x + 1 = 0 \text{with roots} x = [3 \pm \sqrt5]/2;

x 2 − x + 1 = 0 with roots x = [ 1 ± i ∗ 3 ] / 2 ; x^2 - x + 1 = 0 \text{with roots} x = [1 \pm i*\sqrt3]/2;

Hence, x = [ 3 ± 5 ] / 2 x = [3 \pm \sqrt5]/2 are the only real roots.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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