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Question Number 83874 by Power last updated on 07/Mar/20

Commented by niroj last updated on 07/Mar/20

$$\mathrm{xy}({\mathrm{x}}^2-{\mathrm{y}}^2)=24....\left(\mathrm{i}\right)$$$${\mathrm{x}}^2+{\mathrm{y}}^2=10......\left(\mathrm{ii}\right)$$$$\mathrm{multipy}\mathrm{by}\mathrm{xy}\mathrm{in}\left(\mathrm{ii}\right)\mathrm{then}\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)$$$${\mathrm{x}}^3\mathrm{y}-{\mathrm{xy}}^3=24$$$$\frac{{\mathrm{x}}^3\mathrm{y}+{\mathrm{xy}}^3=10\mathrm{xy}}{{2\mathrm{x}}^3\mathrm{y}=10\mathrm{xy}+24}$$$${\mathrm{x}}^3\mathrm{y}=5\mathrm{xy}+12$$$${\mathrm{x}}^3\mathrm{y}-5\mathrm{xy}=12$$$$\mathrm{xy}({\mathrm{x}}^2-5)=12$$$$\mathrm{xy}=\frac{12}{{\mathrm{x}}^2-5}$$$$\mathrm{y}=\frac{12}{{\mathrm{x}}^3-5\mathrm{x}}......\left(\mathrm{iii}\right)$$$$\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\left(\mathrm{ii}\right)$$$${\mathrm{x}}^2+{\mathrm{y}}^2=10$$$${\mathrm{x}}^2+{\left(\frac{12}{{\mathrm{x}}^3-5\mathrm{x}}\right)}^2=10$$$${\mathrm{x}}^2+\frac{144}{{\mathrm{x}}^2{({\mathrm{x}}^2-5)}^2}=10$$$$\mathrm{or},\frac{{\left({\mathrm{x}}^2\right)}^2{({\mathrm{x}}^2-5)}^2+144}{{\mathrm{x}}^2{({\mathrm{x}}^2-5)}^2}=10$$$$\mathrm{or},{\left({\mathrm{x}}^2\right)}^2{({\mathrm{x}}^2-5)}^2+144={10\mathrm{x}}^2{({\mathrm{x}}^2-5)}^2$$$$\mathrm{or},{\left({\mathrm{x}}^2\right)}^2{({\mathrm{x}}^2-5)}^2-{10\mathrm{x}}^2{({\mathrm{x}}^2-5)}^2+144=0$$$${\mathrm{x}}^2-5=\mathrm{t}\Rightarrow {\mathrm{x}}^2=\mathrm{t}+5$$$${(\mathrm{t}+5)}^2{\mathrm{t}}^2-10(\mathrm{t}+5){\mathrm{t}}^2+144=0$$$${\mathrm{t}}^2(\mathrm{t}+5)\{\mathrm{t}+5-10\}+144=0$$$${\mathrm{t}}^2(\mathrm{t}+5)(\mathrm{t}-5)+144=0$$$${\mathrm{t}}^2({\mathrm{t}}^2-25)+144=0$$$${\left({\mathrm{t}}^2\right)}^2-{25\mathrm{t}}^2+144=0$$$${\mathrm{t}}^2=\frac{25\stackrel{-}{+}\sqrt{625-576}}{2}$$$${\mathrm{t}}^2=\frac{25\stackrel{-}{+}\sqrt{49}}{2}=\frac{25\stackrel{-}{+}7}{2}$$$${\mathrm{t}}^2=\frac{25+7}{2}=16(+\mathrm{ve})$$$${\mathrm{t}}^2=\frac{25-7}{2}=9(-\mathrm{ve})$$$${\mathrm{t}}^2=9,16$$$$\mathrm{t}=\stackrel{-}{+}3,\stackrel{-}{+}4$$$$\mathrm{again},\mathrm{if}{\mathrm{t}}^2=9$$$$\mathrm{t}={\mathrm{x}}^2-5$$$${\mathrm{t}}^2={({\mathrm{x}}^2-5)}^2$$$${({\mathrm{x}}^2-5)}^2-9=0$$$$({\mathrm{x}}^2-5+3)({\mathrm{x}}^2-5-3)=0$$$$({\mathrm{x}}^2-2)({\mathrm{x}}^2-8)=0$$$$\mathrm{x}=\underset{-}{+}\sqrt{2},\underset{-}{+}2\sqrt{2}$$$$\mathrm{now}\mathrm{again}\mathrm{if}{\mathrm{t}}^2=16$$$${\mathrm{t}}^2={({\mathrm{x}}^2-5)}^2\Rightarrow 16={({\mathrm{x}}^2-5)}^2$$$$({\mathrm{x}}^2-5)-4^2=0$$$$({\mathrm{x}}^2-5+4)({\mathrm{x}}^2-5-4)=0$$$$({\mathrm{x}}^2-1)({\mathrm{x}}^2-9)=0$$$$\mathrm{x}=\underset{-}{+}1,\underset{-}{+}3$$$$\therefore \mathrm{x}=\underset{-}{+}(1,3,\sqrt{2},2\sqrt{2})$$$$\mathrm{for}\mathrm{y}...\mathrm{also}\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}$$$$\mathrm{in}\mathrm{terms}\mathrm{of}\left(\mathrm{iii}\right).$$$$$$

Answered by john santu last updated on 07/Mar/20

$${\left(\mathrm{xy}\right)}^2=(5+\frac{12}{\mathrm{xy}})(5-\frac{12}{\mathrm{xy}})$$$${\left(\mathrm{xy}\right)}^2=25-\frac{144}{{\left(\mathrm{xy}\right)}^2}$$$$\mathrm{let}{\left(\mathrm{xy}\right)}^2=\mathrm{t}\Rightarrow \mathrm{t}+\frac{144}{\mathrm{t}}-25=0$$$${\mathrm{t}}^2-25\mathrm{t}+144=0$$$$\mathrm{t}=\frac{25\pm \sqrt{625-576}}{2}=\frac{25\pm 7}{2}=16\mathrm{or}9$$$$\mathrm{xy}=\pm 4\mathrm{or}\pm 3\Rightarrow \mathrm{i}\mathrm{think}\mathrm{it}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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