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Question Number 106259 by qwerty111 last updated on 03/Aug/20

Commented by qwerty111 last updated on 03/Aug/20

$$\sqrt{xy}=?$$

Answered by john santu last updated on 03/Aug/20

$$\mathrm{y}\sqrt{\mathrm{x}}=\frac{95}{8}-\mathrm{x}=\frac{95-8\mathrm{x}}{8}$$$$\mathrm{y}=\frac{95-8\mathrm{x}}{8\sqrt{\mathrm{x}}}...\left(1\right)$$$$\mathrm{x}\sqrt{\mathrm{y}}=\frac{93}{8}-\mathrm{y}=\frac{93-8\mathrm{y}}{8}$$$$\mathrm{x}=\frac{93-8\mathrm{y}}{8\sqrt{\mathrm{y}}}...\left(2\right)$$$$\left(1\right):\left(2\right)\to \frac{\mathrm{y}}{\mathrm{x}}=\frac{95-8\mathrm{x}}{8\sqrt{\mathrm{x}}}.\frac{8\sqrt{\mathrm{y}}}{93-8\mathrm{y}}$$$$\frac{\mathrm{y}}{\mathrm{x}}=\sqrt{\frac{\mathrm{y}}{\mathrm{x}}}\left(\frac{95-8\mathrm{x}}{93-8\mathrm{y}}\right)\Rightarrow \frac{95-8\mathrm{x}}{93-8\mathrm{y}}=\frac{\sqrt{\mathrm{y}}}{\sqrt{\mathrm{x}}}$$$$\frac{\mathrm{y}}{\mathrm{x}}={\left(\frac{95-8\mathrm{x}}{93-8\mathrm{y}}\right)}^2...$$$$$$

Answered by Her_Majesty last updated on 03/Aug/20

$$believingthatx,ymustberationalnumbers$$$$andthedenominatorofbothlhsmustbe8$$$$Iguessx=\frac{p^2}{4}\wedge y=\frac{q^2}{4}withp,qnatural$$$$\Rightarrow$$$$\{\begin{array}{l}2p^2+{pq}^2=95\Rightarrow q^2=\frac{95-2p^2}{p}\\ p^{2}q+2q^2=93\Rightarrow p^2=\frac{93-2q^2}{q}\end{array}$$$$now{it}^{\prime }seasytotrybecausep\mid (95-2p^2)and$$$$q\mid (93-2q^2)andbothsquarenaturals$$$$\Rightarrow p=5\wedge q=3$$$$\Rightarrow x=\frac{25}{4}\wedge q=\frac{9}{4}$$$$\Rightarrow \sqrt{xy}=\frac{15}{4}$$

Commented by Her_Majesty last updated on 03/Aug/20

$$whyquestion93???$$

Commented by bemath last updated on 04/Aug/20

$$\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{do}\mathrm{find}\mathrm{p}=5\mathrm{and}\mathrm{q}=3$$

Commented by Her_Majesty last updated on 04/Aug/20

$$bytrying$$

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