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Question Number 112208 by Khalmohmmad last updated on 06/Sep/20

Commented by Dwaipayan Shikari last updated on 06/Sep/20

$$x=9$$$$\left(fromobservation\right)$$$$y=4$$$$$$

Commented by Khalmohmmad last updated on 06/Sep/20

$$\mathrm{solving}\mathrm{method}??$$

Commented by Dwaipayan Shikari last updated on 06/Sep/20

$$x={(7-y)}^2$$$${(7-y)}^2=11-\sqrt{y}(y=a^2)$$$${(7-a^2)}^2=11-a$$$$a^4-14a^2+49=11-a$$$$a^4-14a^2+a+38=0$$$$a=2$$$$y=4$$$$x=3^2=9$$

Answered by bemath last updated on 07/Sep/20

$$\sqrt{\mathrm{x}}=\mathrm{p};\sqrt{\mathrm{y}}=\mathrm{q}$$$$\{\begin{array}{l}\mathrm{p}+{\mathrm{q}}^2=7\\ {\mathrm{p}}^2+\mathrm{q}=11\end{array}\Rightarrow \mathrm{p}=7-{\mathrm{q}}^2$$$$\Rightarrow {(7-{\mathrm{q}}^2)}^2+\mathrm{q}=11,{\mathrm{q}}^4-{14\mathrm{q}}^2+49+\mathrm{q}-11=0$$$${\mathrm{q}}^4-{14\mathrm{q}}^2+\mathrm{q}+38=0$$$$\mathrm{factoring}$$$$(\mathrm{q}-2)({\mathrm{q}}^3+{2\mathrm{q}}^2+4\mathrm{q}-19)=0$$$$\to \{\begin{array}{l}\mathrm{q}=2\to 2=\sqrt{\mathrm{y}},\mathrm{y}=4\\ {\mathrm{q}}^3+{2\mathrm{q}}^2+4\mathrm{q}-19=0\end{array}$$$$$$

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