Let-gt-0-fixed-Solve-for-real-numbers-the-system-x-2-yz-2-y-2-zx-7-2-z-2-xy-5-2-

Question Number 218364 by hardmath last updated on 08/Apr/25

Let 𝛌>0 fixed Solve for real numbers the system: { ((x^2 − yz = λ^2 )),((y^2 − zx = 7λ^2 )),((z^2 − xy = −5λ^2 )) :}

$$\mathrm{Let}\:\:\:\boldsymbol{\lambda}>\mathrm{0}\:\:\:\mathrm{fixed} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{system}:\:\:\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yz}\:=\:\lambda^{\mathrm{2}} }\\{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{zx}\:=\:\mathrm{7}\lambda^{\mathrm{2}} }\\{\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{xy}\:=\:−\mathrm{5}\lambda^{\mathrm{2}} }\end{cases} \\ $$

Answered by vnm last updated on 08/Apr/25

y^2 −x^2 +z(y−x)=6λ^2 (x+y+z)(y−x)=6λ^2 z^2 −y^2 +x(z−y)=−12λ^2 (x+y+z)(z−y)=−12λ^2 x^2 −z^2 +y(x−z)=6λ^2 (x+y+z)(x−z)=6λ^2 y−x=x−z y+z=2x y=(1+t)x z=(1−t)x 3x∙tx=6λ^2 x=λ(√(2/t)) y=λ(1+t)(√(2/t)) z=λ(1−t)(√(2/t)) x^2 −yz=λ^2 λ^2 (2/t)−λ^2 (1−t^2 )(2/t)=λ^2 (2/t)−(1−t^2 )(2/t)=1 2t=1 t=(1/2) x=2λ, y=3λ, z=λ

$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} +{z}\left({y}−{x}\right)=\mathrm{6}\lambda^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)\left({y}−{x}\right)=\mathrm{6}\lambda^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} −{y}^{\mathrm{2}} +{x}\left({z}−{y}\right)=−\mathrm{12}\lambda^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)\left({z}−{y}\right)=−\mathrm{12}\lambda^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{z}^{\mathrm{2}} +{y}\left({x}−{z}\right)=\mathrm{6}\lambda^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)\left({x}−{z}\right)=\mathrm{6}\lambda^{\mathrm{2}} \\ $$$${y}−{x}={x}−{z} \\ $$$${y}+{z}=\mathrm{2}{x} \\ $$$${y}=\left(\mathrm{1}+{t}\right){x} \\ $$$${z}=\left(\mathrm{1}−{t}\right){x} \\ $$$$\mathrm{3}{x}\centerdot{tx}=\mathrm{6}\lambda^{\mathrm{2}} \\ $$$${x}=\lambda\sqrt{\frac{\mathrm{2}}{{t}}} \\ $$$${y}=\lambda\left(\mathrm{1}+{t}\right)\sqrt{\frac{\mathrm{2}}{{t}}} \\ $$$${z}=\lambda\left(\mathrm{1}−{t}\right)\sqrt{\frac{\mathrm{2}}{{t}}} \\ $$$${x}^{\mathrm{2}} −{yz}=\lambda^{\mathrm{2}} \\ $$$$\lambda^{\mathrm{2}} \frac{\mathrm{2}}{{t}}−\lambda^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)\frac{\mathrm{2}}{{t}}=\lambda^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{{t}}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\frac{\mathrm{2}}{{t}}=\mathrm{1} \\ $$$$\mathrm{2}{t}=\mathrm{1} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\mathrm{2}\lambda,\:\:\:{y}=\mathrm{3}\lambda,\:\:\:{z}=\lambda \\ $$

Commented by hardmath last updated on 09/Apr/25

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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