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Question Number 215559 by hardmath last updated on 10/Jan/25

$$\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}=5$$$$\sqrt{\mathrm{x}}\cdot \sqrt{\mathrm{y}}=8$$$$\frac{\sqrt{\mathrm{x}}\mathrm{y}-\mathrm{x}\sqrt{\mathrm{y}}}{\mathrm{y}-\mathrm{x}}=?$$

Answered by Rasheed.Sindhi last updated on 10/Jan/25

$$\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}=5;\sqrt{\mathrm{x}}\sqrt{\mathrm{y}}=8$$$$\frac{\sqrt{\mathrm{x}}\mathrm{y}-\mathrm{x}\sqrt{\mathrm{y}}}{\mathrm{y}-\mathrm{x}}$$$$=\frac{\sqrt{\mathrm{xy}}(\sqrt{\mathrm{y}}-\sqrt{\mathrm{x}})}{{\left(\sqrt{\mathrm{y}}\right)}^2-{\left(\sqrt{\mathrm{x}}\right)}^2}$$$$=\frac{\sqrt{\mathrm{xy}}\cancel{(\sqrt{\mathrm{y}}-\sqrt{\mathrm{x}})}}{\cancel{(\sqrt{\mathrm{y}}-\sqrt{\mathrm{x}})}(\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}})}=\frac{8}{5}$$

Commented by Frix last updated on 11/Jan/25

$$\mathrm{Yes}!$$

Answered by Frix last updated on 11/Jan/25

$$\mathrm{To}\mathrm{find}x,y\mathrm{with}\alpha ,\beta \in \mathbb{R}$$$$A\sqrt{x}+\sqrt{y}=\alpha$$$$B\sqrt{x}\sqrt{y}=\beta$$$$\mathrm{Let}\sqrt{x}=a+b\mathrm{i}\wedge \sqrt{y}=a-b\mathrm{i};b⩾0$$$$A2a=\alpha$$$$B{a}^2+b^2=\beta$$$$=========$$$$Aa=\frac{\alpha }{2}$$$$Bb=\sqrt{\beta -\frac{{\alpha }^2}{4}}$$$$=========$$$$\Rightarrow$$$$\sqrt{x}=\frac{\alpha }{2}+\frac{\sqrt{4\beta -{\alpha }^2}}{2}\mathrm{i}$$$$\sqrt{y}=\frac{\alpha }{2}-\frac{\sqrt{4\beta -{\alpha }^2}}{2}\mathrm{i}$$$$\Rightarrow$$$$x=\frac{{\alpha }^2-2\beta }{2}+\frac{\alpha \sqrt{4\beta -{\alpha }^2}}{2}\mathrm{i}$$$$y=\frac{{\alpha }^2-2\beta }{2}-\frac{\alpha \sqrt{4\beta -{\alpha }^2}}{2}\mathrm{i}$$$$x,y\in \mathbb{R}\iff \beta ⩽\frac{{\alpha }^2}{4}$$$$x=\beta {\mathrm{e}}^{\mathrm{i}{\cos }^{-1}\frac{{\alpha }^2-2\beta }{2\beta }}$$$$y=\beta {\mathrm{e}}^{-\mathrm{i}{\cos }^{-1}\frac{{\alpha }^2-2\beta }{2\beta }}$$

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