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Question Number 11356 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Mar/17

Answered by mrW1 last updated on 21/Mar/17

$$p=\sqrt{x}$$$$q=\sqrt{y}$$$$p^2+q=a$$$$q^2+p=b$$$$q=a-p^2$$$${(a-p^2)}^2+p=b$$$$p^4-2{ap}^2+p+a^2-b=0$$$$similarly$$$$q^4-2{bq}^2+q+b^2-a=0$$$$$$$$working...$$

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