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Question Number 38907 by ajfour last updated on 01/Jul/18

$$\sqrt{c^2+x^2}=1+\frac{c}{x}$$$$howmanycomplexroots?$$

Commented by ajfour last updated on 01/Jul/18

Commented by MJS last updated on 01/Jul/18

$$\sqrt{c^2+x^2}=\frac{c+x}{x}\mathrm{has}\mathrm{only}3\mathrm{roots}.\mathrm{if}\mathrm{you}\mathrm{square}$$$$\mathrm{to}\mathrm{solve}\mathrm{it},\mathrm{you}\mathrm{get}4\mathrm{roots}\mathrm{of}\mathrm{which}\mathrm{one}\mathrm{is}\mathrm{no}$$$$\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{equation}$$

Commented by MrW3 last updated on 01/Jul/18

$${You}^{\prime }rerightsir.Sotheansweris:$$$$dependingonthevalueofc,the$$$$originaleqn.haseithernoneortwo$$$$complexroots.$$

Commented by ajfour last updated on 01/Jul/18

$$Thanksforthetruth,sir.$$

Commented by MJS last updated on 01/Jul/18

$$\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}c=\mathfrak{c}\mathrm{with}$$$$c=\mathfrak{c}\Rightarrow f\left(x\right)=\sqrt{x^2+c^2}-\frac{x+c}{x}=0\mathrm{has}\mathrm{exactly}2\mathrm{real}\mathrm{solutions}$$$$c<\mathfrak{c}\Rightarrow f\left(x\right)=0\mathrm{has}3\mathrm{real}\mathrm{solutions}$$$$c>\mathfrak{c}\Rightarrow f\left(x\right)=0\mathrm{has}1\mathrm{real}\mathrm{and}2\mathrm{complex}\mathrm{solutions}$$$$$$$$x^4+({\mathfrak{c}}^2-1)x^2-2\mathfrak{c}x-{\mathfrak{c}}^2=0$$$${(x-\alpha )}^2(x+\alpha -\beta )(x+\alpha +\beta )=0$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{way}\mathrm{as}\mathrm{several}\mathrm{times}\mathrm{before},\mathrm{in}$$$$\mathrm{this}\mathrm{special}\mathrm{case}\mathrm{we}\mathrm{get}3\mathrm{equations}\mathrm{in}\alpha ,\beta ,\mathfrak{c}$$$$\mathrm{with}\alpha >0,\beta >0,\mathfrak{c}<0\mathrm{and}{\beta }^2\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{real}$$$$\mathrm{solution}\mathrm{of}\mathrm{a}\mathrm{polynome}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{degree}.$$$$x=-\alpha +\beta \mathrm{should}\mathrm{be}\mathrm{the}‘‘{\mathrm{wrong}}^{″}\mathrm{solution}\mathrm{I}\mathrm{guess}$$$$\mathrm{I}\mathrm{might}\mathrm{post}\mathrm{it}\mathrm{later}\mathrm{but}\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{easy}\mathrm{to}$$$$\mathrm{solve}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$$c^2+x^2=1+\frac{2c}{x}+\frac{c^2}{x^2}$$$$c^2+x^2=\frac{x^2+2cx+c^2}{x^2}$$$$x^2{c}^2+x^4=x^2+2cx+c^2$$$$x^4+x^2(c^2-1)-2cx-c^2=0$$$$f\left(x\right)=x^4+0.x^3+(c^2-1)x^2-2cx-c^2$$$$let{c}^2>1usingDescartsruleofsign$$$$onesignchangesoonepositiveroot$$$$f(-x)=x^4+0.(-x^3)+(c^2-1)x^2+2cx-c^2$$$$onesignchangesoonenegdtiveroot$$$$so4-1-1=2complxroot$$$$nowif{c}^2<1$$$$f\left(x\right)=x^4+0.x^3-(1-c^2)x^2+2cx-c^2$$$$2+veroot$$$$f(-x)=x^4-(1-c^2)x^2-2cx-c^2$$$$1(-ve)root$$$$soonecomplexroot$$

Commented by ajfour last updated on 01/Jul/18

$$Thankyouverymuch,SirTanmay.$$

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