Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 172312 by naka3546 last updated on 25/Jun/22

$$x+\frac{2}{x}=2y$$$$y+\frac{2}{y}=2z$$$$z+\frac{2}{z}=2x$$$$(x,y,z)=?$$

Commented by MJS_new last updated on 25/Jun/22

$$\mathrm{there}\mathrm{are}\mathrm{also}\mathrm{solutions}\notin \mathbb{R}$$

Commented by naka3546 last updated on 25/Jun/22

$$\mathrm{How}\mathrm{about}x=y=z=-\sqrt{2}?$$

Commented by infinityaction last updated on 25/Jun/22

$$yessirmistakeinmysolution$$

Commented by Tawa11 last updated on 25/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by infinityaction last updated on 25/Jun/22

$$shareyoursolutionsir$$

Commented by MJS_new last updated on 25/Jun/22

$$\mathrm{same}\mathrm{as}\mathrm{yours},\mathrm{we}\mathrm{have}$$$$1-2p=p(p-2q)=q(q-2)$$$$\Rightarrow$$$$1-2p=p(p-2q)$$$$q(q-2)=p(p-2q)$$$$\mathrm{subtracting}\mathrm{both}\Rightarrow p=-\frac{q^2-2q-1}{2}$$$$\mathrm{inserting}\Rightarrow$$$$q^4-10q^2+8q+1=0$$$$(q-1)(q^3+q^2-9q-1)=0$$$$\Rightarrow$$$$q_1\approx -3.49395921\Rightarrow p_1\approx -9.09783468$$$$q_2\approx -.109916264\Rightarrow p_2\approx .384042943$$$$q_3=1\Rightarrow p_3=1$$$$q_4\approx 2.60387547\Rightarrow p_4\approx -.286208264$$$$\mathrm{we}\mathrm{get}$$$$x_1^2\approx -.104190167$$$$x_2^2\approx -8.62388222$$$$x_3^2=2$$$$x_4^2\approx -1.27192761$$$$\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$

Commented by MJS_new last updated on 25/Jun/22

$$\mathrm{you}\mathrm{did}\mathrm{not}\mathrm{find}\mathrm{out}\mathrm{at}\mathrm{first}\mathrm{because}{p}^2=p=1$$

Commented by MJS_new last updated on 25/Jun/22

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{such}\mathrm{question},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{issue}\mathrm{when}$$$$\mathrm{you}\mathrm{use}q+\mathrm{any}\mathrm{number}$$$$\mathrm{and}\mathrm{my}\mathrm{first}\mathrm{line}\mathrm{is}\mathrm{ok}.\mathrm{we}\mathrm{have}$$$$1-2p=p(p-2q)=q(q-2)$$$$\mathrm{all}\mathrm{are}\mathrm{equal}\Rightarrow$$$$1-2p=p(p-2q)\wedge q(q-2)=p(p-2q)$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{like}a=b=c\Rightarrow a=b\wedge c=b$$

Commented by MJS_new last updated on 25/Jun/22

$$\mathrm{your}\mathrm{version}\mathrm{seems}\mathrm{wrong}$$$$y=px\wedge z=qx$$$$x+\frac{2}{x}=2y\Rightarrow x^2=\frac{2}{2p-1}$$$$y+\frac{2}{y}=2z\Rightarrow x^2=\frac{2}{p(2q-p)}$$$$z+\frac{2}{z}=2x\Rightarrow x^2=\frac{2}{q(2-q)}$$

Commented by infinityaction last updated on 25/Jun/22

$$oksirthanks$$

Commented by infinityaction last updated on 26/Jun/22

$$yessir$$

Answered by manxsol last updated on 29/Jan/23

$$asimpleanswer$$$$x=y\pm \sqrt{y^2-2}$$$$y=z\pm \sqrt{z^2-2}$$$$z=x\pm \sqrt{x^2-2}$$$$sumo$$$$\pm \sqrt{y^2-2}\pm \sqrt{z^2-2}\pm \sqrt{x^2-2}=0$$$$unaposibilidad$$$$+++=0---=0$$$$y=\pm \sqrt{2}$$$$x=\pm \sqrt{2}$$$$z=\pm \sqrt{2}$$$$andikeepthinking$$$$otrasposibilidades$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com