solve-x-4-4x-1-

Question Number 188660 by mr W last updated on 04/Mar/23

s o l v e x^{4} + 4 x = 1

Answered by aba last updated on 04/Mar/23

∙ x^{4} = {(x^{2} + 1)}^{2} - {2 x}^{2} - 1

x^{4} + 4 x - 1 = 0 \Rightarrow {(x^{2} + 1)}^{2} - {2 x}^{2} - 1 + 4 x - 1 = 0

\Rightarrow {(x^{2} - 1)}^{2} - 2 (x^{2} - 2 x + 1) = 0

\Rightarrow {(x^{2} + 1)}^{2} - 2 {(x - 1)}^{2} = 0

\Rightarrow x^{2} + 1 = \pm (x - 1) \sqrt{2}

\Rightarrow x^{2} - \sqrt{2} x + (1 + \sqrt{2}) = 0 \lor x^{2} + \sqrt{2} x + (1 - \sqrt{2}) = 0

x^{2} - \sqrt{2} x + (1 + \sqrt{2}) = 0

Δ = - 2 - 4 \sqrt{2} = - 2 (2 \sqrt{2} - 1) < 0

x^{2} + \sqrt{2} x + (1 - \sqrt{2}) = 0

Δ = - 2 + 4 \sqrt{2} = 2 (2 \sqrt{2} - 1) > 0

x = \frac{- \sqrt{2} \pm \sqrt{2 (2 \sqrt{2} - 1)}}{2}

Commented by BaliramKumar last updated on 04/Mar/23

a n s w e r i s r i g h t b u t

l a s t 3 r d a n d 4 t h s t e p e r r o r (t a k e \pm v a l u e)

Answered by manxsol last updated on 05/Mar/23

x^{4} + 2 x^{2} + 1 - 2 x^{2} + 4 x - 1 = 1

{(x^{2} + 1)}^{2} - 2 (x^{2} - 2 x + 1) = 0

{(x^{2} + 1)}^{2} - {(\sqrt{2} x - \sqrt{2})}^{2} = 0

(x^{2} + \sqrt{2} x + 1 - \sqrt{2}) (x^{2} - \sqrt{2} x + 1 + \sqrt{2}) = 0

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