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Question Number 217732 by ArshadS last updated on 19/Mar/25

{\begin{cases} x + y = x y \\ x^{2} + y^{2} = 25 \end{cases}; x^{4} + y^{4} = ?

Commented by Ghisom last updated on 19/Mar/25

x^{4} + y^{4} = 571 \pm 4 \sqrt{26}

Answered by Wuji last updated on 19/Mar/25

S = x + y, P = x y

{(x y)}^{2} = P^{2} = S^{2}

x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 {(x y)}^{2}

x^{4} + y^{4} = {(25)}^{2} - 2 S^{2} = 625 - 2 S^{2}

x^{2} + y^{2} = S^{2} - 2 P \Rightarrow S^{2} - 2 S = 25

S^{2} = 2 S + 25

\Rightarrow x^{4} + y^{4} = 625 - 2 (2 S + 25) = 625 - 4 S - 50

x^{4} + y^{4} = 575 - 4 S

f r o m S^{2} - 2 S - 25 = 0

S = \frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \times (- 25)}}{2} = \frac{2 \pm \sqrt{4 + 100}}{2} = \frac{2 \pm \sqrt{104}}{2}

S = \frac{2 \pm 2 \sqrt{26}}{2} = 1 \pm \sqrt{26}

S = 1 + \sqrt{26} S > 0

\Rightarrow x^{4} + y^{4} = 575 - 4 (1 + \sqrt{26})

x^{4} + y^{4} = 575 - 4 - 4 \sqrt{26}) = 571 - 4 \sqrt{26}

∴ x^{4} + y^{4} = 571 - 4 \sqrt{26}

Commented by ArshadS last updated on 22/Mar/25

T h a n k s a l o t!

Answered by Rasheed.Sindhi last updated on 19/Mar/25

x^{2} + y^{2} = 25

\Rightarrow {(x + y)}^{2} - 2 x y = 25

\Rightarrow {(x + y)}^{2} - 2 (x + y) = 25

\Rightarrow {(x + y)}^{2} - 2 (x + y) - 25 = 0

x + y = x y = \frac{2 \pm \sqrt{4 + 100}}{2} = 1 \pm \sqrt{26}

{(x^{2} + y^{2})}^{2} = 25^{2}

x^{4} + y^{4} + 2 {(x y)}^{2} = 625

x^{4} + y^{4} = 625 - 2 {(1 \pm \sqrt{26})}^{2}

= 625 - 2 (1 + 26 \pm 2 \sqrt{26})

= 625 - 54 \mp 4 \sqrt{26}

= 571 \mp 4 \sqrt{26}

= 571 \pm 4 \sqrt{26}

Commented by ArshadS last updated on 22/Mar/25

T h a n k s s i r!

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