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Question Number 14742 by Mr Chheang Chantria last updated on 04/Jun/17

$$\mathit{x}+\mathit{y}=\frac{1}{8}$$$$\sqrt[4]{\mathit{x}}+\sqrt[4]{\mathit{y}}=1$$$$\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}$$

Commented by Tinkutara last updated on 04/Jun/17

$$\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{verify}\mathrm{that}\frac{1}{16}+\frac{1}{16}=\frac{1}{8}$$$$\mathrm{and}\sqrt[4]{\frac{1}{16}}+\sqrt[4]{\frac{1}{16}}=1$$$$\mathrm{Hence}x=y=\frac{1}{16}$$

Commented by RasheedSoomro last updated on 04/Jun/17

$$\mathrm{Let}\sqrt[4]{\mathit{x}}=\mathrm{p}\mathrm{and}\sqrt[4]{y}=\mathrm{q}$$$$\Rightarrow \{\begin{array}{l}{\mathrm{p}}^4+{\mathrm{q}}^4=1/8\\ \mathrm{p}+\mathrm{q}=1\end{array}$$$$\Rightarrow \frac{{\mathrm{p}}^4+{\mathrm{q}}^4}{\mathrm{p}+\mathrm{q}}=\frac{1}{8}\wedge ({\mathrm{p}}^4+{\mathrm{q}}^4)(\mathrm{p}+\mathrm{q})=\frac{1}{8}$$$$\Rightarrow \frac{{\mathrm{p}}^4+{\mathrm{q}}^4}{\mathrm{p}+\mathrm{q}}-({\mathrm{p}}^4+{\mathrm{q}}^4)(\mathrm{p}+\mathrm{q})=0$$$$\Rightarrow ({\mathrm{p}}^4+{\mathrm{q}}^4)\{\frac{1}{\mathrm{p}+\mathrm{q}}-(\mathrm{p}+\mathrm{q})\}=0$$$${\mathrm{p}}^4+{\mathrm{q}}^4=0\vee \mathrm{p}+\mathrm{q}-\frac{1}{\mathrm{p}+\mathrm{q}}=0$$$${\mathrm{p}}^4=-{\mathrm{q}}^4\vee \mathrm{p}+\mathrm{q}=\frac{1}{\mathrm{p}+\mathrm{q}}$$$$\mathrm{x}=-\mathrm{y}\vee {(\mathrm{p}+\mathrm{q})}^2=1$$$$\mathrm{p}+\mathrm{q}=\pm 1$$$$\mathrm{p}+\mathrm{q}=1\mathrm{is}\mathrm{what}\mathrm{we}\mathrm{have}\mathrm{been}\mathrm{given}!\mathrm{hahahaa}..$$$$\mathrm{I}\mathrm{reached}\mathrm{at}\mathrm{the}\mathrm{starting}\mathrm{point}\mathrm{again}!!!$$$$\mathrm{p}+\mathrm{q}=-1\mathrm{is}\mathrm{extraneous}(?)$$$$$$$$\mathrm{Oh}\mathrm{no}{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{gained}\mathrm{something}\mathrm{x}=-\mathrm{y}$$$$\mathrm{But}\mathrm{this}\mathrm{is}\mathrm{again}\mathrm{in}\mathrm{contradiction}\mathrm{to}\mathrm{the}\mathrm{given}$$$$(\mathrm{x}+\mathrm{y}=1/8)$$$$!!!$$$$\mathrm{Continue}$$

Commented by prakash jain last updated on 04/Jun/17

$$\sqrt[4]{x}=u$$$$\sqrt[4]{y}=v$$$$u^4+v^4=\frac{1}{8}$$$$u+v=1$$$$u^2+v^2+2uv=1$$$$u^2+v^2=1-2uv$$$$u^4+v^4+2u^2{v}^2={(1-2uv)}^2$$$$\frac{1}{8}+2u^2{v}^2=1-4uv+4u^2{v}^2$$$$1+16v^2{u}^2=8-32uv+32u^2{v}^2$$$$16u^2{v}^2-32uv+7=0$$$$(4uv-7)(4uv-1)=0$$$$uv=\frac{1}{4}oruv=\frac{7}{4}$$$$uv=\frac{1}{4}$$$$u+v=1$$$${(u-v)}^2={(u+v)}^2-4uv=1-1=0$$$$u=v=\frac{1}{2}\Rightarrow x=\frac{1}{16},y=\frac{1}{16}$$$$uv=\frac{7}{4}$$$${(u-v)}^2=1-7=-6$$$$u-v=\pm i\sqrt{6}$$$$u=\frac{1\pm i\sqrt{6}}{2},v=\frac{1\mp i\sqrt{6}}{2}$$$$x=\frac{1\pm 20i\sqrt{6}}{16},y=\frac{1\mp 20i\sqrt{6}}{16}$$

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