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If-x-y-and-z-are-numbers-integers-different-of-0-49x-7y-z-0-25x-5y-z-0-find-y-2-4xz-




Question Number 112461 by dw last updated on 08/Sep/20
If   x, y and z  are numbers  integers different of   0   { ((49x+7y+z=0)),((25x−5y+z=0)) :}         find (√(y^2 −4xz))
$${If}\:\:\:{x},\:{y}\:{and}\:{z}\:\:{are}\:{numbers}\:\:{integers}\:{different}\:{of}\:\:\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{49}{x}+\mathrm{7}{y}+{z}=\mathrm{0}}\\{\mathrm{25}{x}−\mathrm{5}{y}+{z}=\mathrm{0}}\end{cases}\:\:\:\:\:\:\:\:\:{find}\:\sqrt{{y}^{\mathrm{2}} −\mathrm{4}{xz}} \\ $$
Answered by floor(10²Eta[1]) last updated on 08/Sep/20
z=5y−25x  49x+7y+5y−25x=24x+12y=0⇒y=−2x  ⇒z=−35x  ⇒x=−k, y=2k, z=35k ∀ k∈Z  (√(y^2 −4xz))=(√(4k^2 +140k^2 ))=12∣k∣
$$\mathrm{z}=\mathrm{5y}−\mathrm{25x} \\ $$$$\mathrm{49x}+\mathrm{7y}+\mathrm{5y}−\mathrm{25x}=\mathrm{24x}+\mathrm{12y}=\mathrm{0}\Rightarrow\mathrm{y}=−\mathrm{2x} \\ $$$$\Rightarrow\mathrm{z}=−\mathrm{35x} \\ $$$$\Rightarrow\mathrm{x}=−\mathrm{k},\:\mathrm{y}=\mathrm{2k},\:\mathrm{z}=\mathrm{35k}\:\forall\:\mathrm{k}\in\mathbb{Z} \\ $$$$\sqrt{\mathrm{y}^{\mathrm{2}} −\mathrm{4xz}}=\sqrt{\mathrm{4k}^{\mathrm{2}} +\mathrm{140k}^{\mathrm{2}} }=\mathrm{12}\mid\mathrm{k}\mid \\ $$
Commented by dw last updated on 08/Sep/20
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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