Solving a Nice Rational Equation with Substitution | Math Olympiad Preparation

[@RealQinnMalloryu4]

(x^2)^2: ➖ (4x)^2/(x^2+x^2 ➖ )(4+4 ➖)= (x^4 ➖ 4x^4)/(x^4+8)= 4x/8 +((x^2)^2 ➖ (4)^2/(x^4)^2 ➖ (16)^2=(x^4 ➖ 16)/(x^16 ➖ 256) =12/240 (4x/8+12/240)= 16x/248=4^4x/10^20^6^8 2^22^2^x/2^5^5^4^3^2^2^3 1^1^1^1^x/1^1^12^2^3^1^1 /1^2 3^1 23 (x ➖ 3x+2)

[@RealQinnMalloryu4]

3(x ➖ 16)^2/x^4 ➖ 16 3(x ➖ 2^2)^2/x^2^2 ➖ 4^4= 3(x ➖ 1^1)^1/x^1^1 ➖ 2^2^2^2 3/1^1^1^2 32 (x ➖ 3x+2).

[@RajeshKumar-wu7ox]

X=2,(-1+2√13) ,(-1-2√13)

[@ashokdubey8415]

x can't be 2 as (x^4-16)=0 in denominator which is absurd

[@RajeshKumar-wu7ox]

Yes sir