Japanese Olympiad Mathematics | This is beautifully solved.

[@brianwade4179]

4*sqrt(x) + x = 8
This is a quadratic in sqrt(x).
Write it as x + 4*sqrt(x) - 8 = 0
Use quadratic formula: a=1, b=4, c=-8
sqrt(x) = (-4 +- sqrt((-4)^2-4(1)(-8))) / 2(1)
aka sqrt(x) = -2 +- sqrt(48)/2
Simplify to sqrt(x) = -2+-2*sqrt(3)
To find the x values, square the two sqrt(x) values:
x1 = (-2+2*sqrt(3))^2 = 4 - 4*sqrt(3) -4*sqrt(3) + 12 = 16 - 8*sqrt(3)
x2 = (-2-2*sqrt(3))^2 = 4 + 4*sqrt(3) + 4*sqrt(3) + 12 = 16 + 8*sqrt(3)

[@OFFICIAL-PO-j2v]

I've been watching a lot of your videos lately, keep up the good work

[@PhilCoolMath]

@@OFFICIAL-PO-j2v thanks 👍

[@kenkramer9015]

A suggested 2-minute solution:
(4 sqrt x)^2 = (8 - x)^2
16x = x^2 - 16x + 64
x^2 -32x + 64 = 0 (Then complete the square)
x^2 - 32x + 256 = -64 + 256
Sqrt(x - 16)^2 = Sqrt192
x - 16 = +\- Sqrt192
x = 16 +\- 8Sqrt3 👍😊