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 194363 by sonukgindia last updated on 05/Jul/23

Commented by Frix last updated on 05/Jul/23

$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{solution}.$$$$\sqrt{z_1+\sqrt{z_2}}+\sqrt{z_1-\sqrt{z_2}}=r$$$$\mathrm{Solving}\mathrm{by}2\mathrm{times}[\mathrm{squaring}(\mathrm{introduces}$$$$\mathrm{f}\mathrm{alse}\mathrm{solutions})\mathrm{and}\mathrm{transforming}]\mathrm{leads}\mathrm{to}:$$$$z_1=\frac{r^4+4z_2+r^4}{4r^2}$$$$z_1=a+b\mathrm{i}\wedge z_2=c+d\mathrm{i}\mathrm{with}a,b,c,d\in \mathbb{R}$$$$a+b\mathrm{i}=\frac{4c+r^4}{4r^2}+\frac{d}{r^2}\mathrm{i}$$$$\{\begin{array}{l}a=\frac{4c+r^4}{4r^2}\\ b=\frac{d}{r^2}\end{array}$$$$\mathrm{We}\mathrm{have}r=2\wedge c=a+17$$$$\Rightarrow a=7\wedge d=4b\Rightarrow$$$$f\left(b\right)=\sqrt{7+b\mathrm{i}+2\sqrt{6+b\mathrm{i}}}+\sqrt{7+b\mathrm{i}-2\sqrt{6+b\mathrm{i}}}$$$$\mathrm{The}\mathrm{real}\mathrm{part}\mathrm{of}f\left(b\right)\mathrm{is}⩾2\sqrt{6}$$$$\mathrm{The}\mathrm{imaginary}\mathrm{part}\mathrm{of}f\left(b\right)\mathrm{has}\mathrm{only}\mathrm{one}$$$$\mathrm{zero}\mathrm{at}b=0$$$$\Rightarrow f\left(0\right)=2\sqrt{6}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{real}\mathrm{point}\mathrm{of}f\left(b\right)$$

Answered by MacX last updated on 05/Jul/23

$$\sqrt{z+\sqrt{z+17}}+\sqrt{z-\sqrt{z+17}}=2$$$$\Rightarrow z+\sqrt{z+17}+z-\sqrt{z+17}+2\sqrt{z^2-(z+17)}=4$$$$\Rightarrow 2z+2\sqrt{z^2-z-17}=4$$$$\Rightarrow 2\sqrt{z^2-z-17}=4-2z=2(2-z)$$$$\Rightarrow z^2-z-17=z^2-4z+4$$$$\Rightarrow -z-17=-4z+4$$$$\Rightarrow 3z=21$$$$\Rightarrow z=7$$$$S_{\mathbb{C}}=\left\{7\right\}$$

Commented by Frix last updated on 05/Jul/23

$$\sqrt{7+\sqrt{7+17}}+\sqrt{7-\sqrt{7+17}}=\sqrt{7+\sqrt{24}}+\sqrt{7-\sqrt{24}}=$$$$=(1+\sqrt{6})+(-1+\sqrt{6})=2\sqrt{6}\ne 2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com