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Question Number 112461 by dw last updated on 08/Sep/20

$$Ifx,yandzarenumbersintegersdifferentof0$$$$\{\begin{array}{l}49x+7y+z=0\\ 25x-5y+z=0\end{array}find\sqrt{y^2-4xz}$$

Answered by floor(10²Eta[1]) last updated on 08/Sep/20

$$\mathrm{z}=5\mathrm{y}-25\mathrm{x}$$$$49\mathrm{x}+7\mathrm{y}+5\mathrm{y}-25\mathrm{x}=24\mathrm{x}+12\mathrm{y}=0\Rightarrow \mathrm{y}=-2\mathrm{x}$$$$\Rightarrow \mathrm{z}=-35\mathrm{x}$$$$\Rightarrow \mathrm{x}=-\mathrm{k},\mathrm{y}=2\mathrm{k},\mathrm{z}=35\mathrm{k}\mathrm{\forall }\mathrm{k}\in \mathbb{Z}$$$$\sqrt{{\mathrm{y}}^2-4\mathrm{xz}}=\sqrt{{4\mathrm{k}}^2+{140\mathrm{k}}^2}=12\mid \mathrm{k}\mid$$

Commented by dw last updated on 08/Sep/20

$$ThankyouSir$$

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