Math Olympiad | A Nice Rational Equation | VIJAY Maths

[@RyanLewis-Johnson-wq6xs]

(x^7+x^5+x^3)/(x^6+x^5+x^4)=81/27 x=2 ± Sqrt[3]

[@souzasilva5471]

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?)

[@ghaidabarghouthi1403]

You have done extra steps, just divide 81/27=3 from the beginning

[@vijaymaths5483]

Yes ,I know thanks for sharing 👍 your valuable feedback 🌺

[@kelkarbharat9161]

It is surprising that 81/27 is continued throught out instead of writing it as 3. I don't understand the reason!

[@uttiyamajumdar8928]

(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

[@RealQinnMalloryu4]

{x^7+x^7 ➖ }+{x^5+x^5 ➖ }+{x^3+x^3 ➖ }/(x^6+x^6 ➖}+(x^5+x^5 ➖}+{x^4+x^4 ➖}= 30x^6/30x^6=1x1 (x ➖ 1x+1). 9^9/3^3 3^2^3^2/1^1 1^1^3^1 (x ➖ 3x+1)