solve-x-2-5-2-x-5-

Question Number 178251 by mr W last updated on 14/Oct/22

s o l v e

{(x^{2} - 5)}^{2} - x = 5

Answered by BaliramKumar last updated on 14/Oct/22

{(x^{2} - 5)}^{2} - x = 5

x^{2} - 5 = \sqrt{5 + x}

x^{2} = 5 + \sqrt{5 + x}

x = \sqrt{5 + \sqrt{5 + x}}

x = \sqrt{5 + \sqrt{5 + \sqrt{5 + \dots \dots \dots . . \infty}}}

x = \sqrt{5 + x}

x^{2} - x - 5 = 0

x = \frac{1 \pm \sqrt{21}}{2}

Commented by mr W last updated on 14/Oct/22

t h a n k s!

Answered by Frix last updated on 14/Oct/22

{(x^{2} - 5)}^{2} - x - 5 = 0

x^{4} - 10 x^{2} - x + 20 = 0

(x^{2} + x - 4) (x^{2} - x - 5) = 0

x = - \frac{1 \pm \sqrt{17}}{2} \lor x = \frac{1 \pm \sqrt{21}}{2}

Commented by mr W last updated on 14/Oct/22

t h a n k s!

Answered by mr W last updated on 14/Oct/22

{(x^{2} - 5)}^{2} - x^{2} = 5 + x - x^{2}

(x^{2} - 5 + x) (x^{2} - 5 - x) = - (x^{2} - 5 - x)

(x^{2} - 4 + x) (x^{2} - 5 - x) = 0

\Rightarrow x^{2} - 4 + x = 0 \Rightarrow x = \frac{- 1 \pm \sqrt{17}}{2}

\Rightarrow x^{2} - 5 - x = 0 \Rightarrow x = \frac{1 \pm \sqrt{21}}{2}

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