✍️ Posted by Ramanujan Algebra Problem

[@ashokpatwardhan3572]

Good

[@raghvendrasingh1289]

eliminating b
√a+(14-a)^2 = 28
let √a = x
x+196+x^4 - 28 x^2-28= 0
x^4 - 28 x^2+x+168 = 0
x = 3 by RRT
a = 9
from second equation
a+√b = 14
√b = 5
b = 25

[@RealQinnMalloryu4]

(14)+(14)=28 (2^2)+(2^2) (1^1)+(1^2) (1^2) (b ➖ 2a+1) (7)+(7)=14 (3^4)+(3^4) (3^2^2)+(3^2^2).(1^1^1)+(3^1^2) (3^2) (b ➖ 3a+2).