Question Number 214457 by hardmath last updated on 09/Dec/24
$$\mathrm{If}\:\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{3z}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} }\:=\:\mathrm{14}\:\:\:\:\:\mathrm{find}:\:\:\frac{\mathrm{x}\:+\:\mathrm{y}}{\mathrm{z}}\:=\:? \\ $$
Commented by Ghisom last updated on 09/Dec/24
$$\mathrm{1} \\ $$
Commented by hardmath last updated on 09/Dec/24
$$\mathrm{how},\:\mathrm{solution},\:\mathrm{if}\:\mathrm{please} \\ $$
Answered by Frix last updated on 09/Dec/24
$$\left({x}+\mathrm{2}{y}+\mathrm{3}{z}\right)^{\mathrm{2}} =\mathrm{14}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$\mathrm{13}{x}^{\mathrm{2}} +\mathrm{10}{y}^{\mathrm{2}} +\mathrm{5}{z}^{\mathrm{2}} −\mathrm{4}{xy}−\mathrm{6}{xz}−\mathrm{12}{yz}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{3}\left({x}+\mathrm{2}{y}\right)}{\mathrm{5}}\pm\frac{\sqrt{\mathrm{14}}\left(\mathrm{2}{x}−{y}\right)}{\mathrm{5}}\mathrm{i} \\ $$$$\mathrm{Solution}\:\mathrm{only}\:\mathrm{unique}\:\mathrm{if}\:{z}\in\mathbb{R}\:\Rightarrow \\ $$$${y}=\mathrm{2}{x}\wedge{z}=\mathrm{3}{x} \\ $$$$\Rightarrow \\ $$$$\frac{{x}+{y}}{{z}}=\mathrm{1} \\ $$
Commented by hardmath last updated on 09/Dec/24
$$ \\ $$Is there a shorter succinct solution to this?
Answered by mr W last updated on 09/Dec/24
![say ((x+y)/z)=(1/k), then z=k(x+y) [x+2y+3k(x+y)]^2 =14x^2 +14y^2 +14k^2 (x+y)^2 (5k^2 −6k+13)x^2 +(5k^2 −12k+10)y^2 +2(5k^2 −9k−2)xy=0 Δ=(5k^2 −9k−2)^2 x^2 −(5k^2 −12k+10)(5k^2 −6k+13)x^2 ≥0 (5k^2 −9k−2)^2 −(5k^2 −9k−3k+10)(5k^2 −9k+3k+13)≥0 −126(k−1)^2 ≥0 the only solution for this inequality is k−1=0, i.e. k=1. ⇒((x+y)/z)=(1/k)=(1/1)=1 ✓](https://www.tinkutara.com/question/Q214481.png)
$${say}\:\frac{{x}+{y}}{{z}}=\frac{\mathrm{1}}{{k}},\:{then}\:{z}={k}\left({x}+{y}\right) \\ $$$$\left[{x}+\mathrm{2}{y}+\mathrm{3}{k}\left({x}+{y}\right)\right]^{\mathrm{2}} =\mathrm{14}{x}^{\mathrm{2}} +\mathrm{14}{y}^{\mathrm{2}} +\mathrm{14}{k}^{\mathrm{2}} \left({x}+{y}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{13}\right){x}^{\mathrm{2}} +\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{12}{k}+\mathrm{10}\right){y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{9}{k}−\mathrm{2}\right){xy}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{9}{k}−\mathrm{2}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{12}{k}+\mathrm{10}\right)\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{13}\right){x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{9}{k}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{9}{k}−\mathrm{3}{k}+\mathrm{10}\right)\left(\mathrm{5}{k}^{\mathrm{2}} −\mathrm{9}{k}+\mathrm{3}{k}+\mathrm{13}\right)\geqslant\mathrm{0} \\ $$$$−\mathrm{126}\left({k}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${the}\:{only}\:{solution}\:{for}\:{this}\:{inequality} \\ $$$${is}\:{k}−\mathrm{1}=\mathrm{0},\:{i}.{e}.\:{k}=\mathrm{1}. \\ $$$$\Rightarrow\frac{{x}+{y}}{{z}}=\frac{\mathrm{1}}{{k}}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1}\:\checkmark \\ $$
Commented by hardmath last updated on 10/Dec/24
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$