MathDB
(x+1)^2-5(x+1) \sqrt{x}+4x=0 (II Soros Olympiad 1995-96 R2 9.1)

Source:

June 3, 2024
algebra

Problem Statement

Solve the equation (x+1)25(x+1)x+4x=0(x+1)^2-5(x+1) \sqrt{x}+4x=0