Solving x^4−2x^2−x+1=0 for x

Nayeli Osborne

Nayeli Osborne

Answered question

2022-10-12

Solving x 4 2 x 2 x + 1 = 0 for x

Answer & Explanation

imperiablogyy

imperiablogyy

Beginner2022-10-13Added 13 answers

For:
a x 4 + c x 2 + d x + e = 0
The solutions are:
x 1 , 2 = S ± 1 2 4 S 2 + d 2 c S S a
x 3 , 4 = + S ± 1 2 4 S 2 d + 2 c S S a
Where:
S = 1 2 3 a 2 c + Q + Δ 0 Q
Q = 1 2 3 Δ 1 + Δ 1 2 4 Δ 0 3 3
Δ 0 = c 2 + 12 a e ,     Δ 1 = 2 c 3 + 27 a d 2 72 a c e
In your case, the two real solutions are x 3 , 4 , and you can verify that:
Q = 1 2 ( 155 + 3 849 ) 3
S = 1 2 3 4 + Q + 16 Q
x = S ± 1 2 3 a 4 + 1 S 4 S 2 0.5249 , 1.4902

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