Math Olympiad Algebra Equation | Nice Radical Simplification Problem | Find X in this Question ?

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[richardleveson]

Thank you - this was worth all the hard work and temporarily replacing 7 with variable n avoids having to solve a fourth degree equation, replacing it with two quadratics. I would not have thought of doing that! Clever solution.

[ivanhuertas]

.thanks for your videos one curiosity if you make the graphic of the funcion when you equal the equation to cero you get just one value x=16 but if you use substitution therefore getting a quartic equation its graphic gives the 2 rational values and the 2 no rational values

[walterwen]

Math Olympiad Algebra Equation: √x + √(⁴√x + 7) = 7; x = ?
7 > √x > √(⁴√x + 7) > 0, Let: y = ⁴√x > 0, x = y⁴; y² + √(y + 7) = 7
y + 7 = (7 – y²)² = y⁴ – 14y² + 49, y⁴ – 14y² – y + 42 = 0
(y⁴ – 14y² + 45) – y – 3 = (y² – 9)(y² – 5) – (y + 3) = 0
(y + 3)[(y – 3)(y² – 5) – 1] = (y + 3)(y³ – 3y² – 5y + 14); y + 3 > 0
y³ – 3y² – 5y + 14 = y²(y – 2) – (y² + 5y – 14) = y²(y – 2) – (y + 7)(y – 2) = 0
(y – 2)(y² – y – 7) = 0; y – 2 = 0 or y² – y – 7 = 0
y = 2 = ⁴√x, x = 2⁴ = 16 or y = (1 + √29)/2 > 3, √x = y² > 7; Rejected
Answer check:
x = 16 = 2⁴; √x + √(⁴√x + 7) = 2² + √(2 + 7) = 4 + 3 = 7; Confirmed
Final answer:
x = 16

[alishernizamov]

x>0 and 7-sqrt(x)>0, sqrt(x)<7, x<49; 0<x<49

[davidshen]

A=X^（1/4）， Sqrt（X）=A^2， A^2+Sqrt（A+7）=7， B=Sqrt（A+7）， A^2+B=7， B^2=A+7， A^2-B^2+B+A=0，（A-B+1）（A+B）=0， A+B！=0， A-B+1=0， A^2+A-6=0， A=3，