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Question Number 42316 by Raj Singh last updated on 23/Aug/18

Commented by Raj Singh last updated on 23/Aug/18

$$howx=4y=9$$

Commented by maxmathsup by imad last updated on 23/Aug/18

$$letput\sqrt{x}=uand\sqrt{y}=v\Rightarrow u+v^2=11and{u}^2+v=7\Rightarrow$$$$u=11-v^{2}and{(11-v^2)}^2+v=7\Rightarrow v^4-22v^2+121+v-7=0\Rightarrow$$$$v^4-22v^2+v+114=0thesolutionofthisequationare$$$$v_1\sim -3,7793\left(real\right)$$$$v_2\sim 3,5844\left(real\right)$$$$v_3=3\left(real\right)$$$$v_4\sim -2,8051\left(real\right)butvmustbe+ve\Rightarrow v=v_{2}or{v}_3$$$$v=3,7793\Rightarrow y={(3,7793)}^{2}andu=11-{(3,7793)}^2=...$$$$v=3\Rightarrow y=9andu=11-3^2=2=\sqrt{x}\Rightarrow x=4.$$

Answered by behi83417@gmail.com last updated on 23/Aug/18

$$(\sqrt{x}-2)+(\sqrt{y}-3)(\sqrt{y}+3)=0$$$$(\sqrt{x}-2)(\sqrt{x}+2)+(\sqrt{y}-3)=0$$$$(\sqrt{x}-2)-(\sqrt{x}-2)(\sqrt{x}+2)(\sqrt{y}+3)=0$$$$(\sqrt{x}-2)(1-(\sqrt{x}+2)(\sqrt{y}+3))=0$$$$\Rightarrow \sqrt{x}-2=0\Rightarrow x=4$$$$x+\sqrt{y}=7\Rightarrow \sqrt{y}=7-4=3\Rightarrow y=9.◼$$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Aug/18

$$bahdarunexcellent....$$

Commented by Joel578 last updated on 23/Aug/18

$$\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{imagine}\mathrm{how}\mathrm{you}\mathrm{get}\mathrm{this}\mathrm{idea}$$

Commented by maxmathsup by imad last updated on 23/Aug/18

$$ithinksirbehihaveplayedamatchcriketwiththisequation...$$

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