Porque as raízes complexas não servem? (x + 1/x = -1) se não foi dado o conjunto prè especificado? (Why are complex roots no good? (x + 1/x = -1) if the pre-specified set was not given?)
@ghaidabarghouthi14032 ай бұрын
You have done extra steps, just divide 81/27=3 from the beginning
@vijaymaths54832 ай бұрын
Yes ,I know thanks for sharing 👍 your valuable feedback 🌺
@kelkarbharat9161Ай бұрын
It is surprising that 81/27 is continued throught out instead of writing it as 3. I don't understand the reason!
@uttiyamajumdar8928Ай бұрын
(x^7+x^5+x^3)/(x^6+x^5+x^4 )=81/27 ⇒x^3/x^4 .(x^4+x^2+1)/(x^2+x+1)=81/27 ⇒1/x.(x^4+x^2+1)/(x^2+x+1)=3^4/3^3 ……(1) Now,x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1) Therefore, (1)⇒1/x.(x^2-x+1)(x^2+x+1)/((x^2+x+1) )=3 ⇒(x^2-x+1)(x^2+x+1)=3x(x^2+x+1) ⇒(x^2+x+1)(x^2-x+1-3x)=0 ⇒(x^2+x+1)(x^2-4x+1)=0 ⇒Either,x^2+x+1=0 Or, x^2-4x+1=0 But,x^2+x+1≠0 as then the given equation does not hold good.Therefore, x^2-4x+1=0 x=(4±√(16-4))/2 =(4±2√3)/2 =2±√3