Japanese | Can You Solve This?? | A Nice Nath Olympiad Algebra Problem | No Calculator Allowed

[@PeterArkadiev]

Since both equations are symmetrical, the two "different" solutions are just notational variants

[@guyhoghton399]

Let _(t - √x)(t - √y) = 0_ ... ①
⇒ _t² - (√x + √y)t + √(xy) = 0_
⇒ _t² - 10t + 10 = 0_
⇒ _t = 5 ± √15_
⇒ _( t - (5 + √15) )( t - (5 - √15) ) = 0_
∴ by comparison with ①:
_√x = 5 + √15, √y = 5 - √15_ up to symmetry
⇒ *_x = 40 + 10√15, y = 40 - 10√15_*

[@stpat7614]

I did it a different and (unfortunately) longer way.
sqrt(x) + sqrt(y) = 10
Since sqrt(x) >= 0, then x >= 0
Since sqrt(y) >= 0, then y >= 0
sqrt(xy) = 10
Since sqrt(xy) > 0, then x > 0 and y > 0
[sqrt(x) + sqrt(y)]^2 = 10^2
[sqrt(x)]^2 + [sqrt(y)]^2 + 2*sqrt(x)*sqrt(y) = 100
x + y + 2*sqrt(xy) = 100
x + y + 2*sqrt(xy) - x - y = 100 - x - y
2*sqrt(xy) = 100 - x - y
2*sqrt(xy)/2 = [100 - x - y]/2
sqrt(xy) = [100 - x - y]/2
[100 - x - y]/2 = sqrt(xy)
Since sqrt(xy) = 10, then
[100 - x - y]/2 = 10
[100 - x - y]/2*2 = 10*2
100 - x - y = 20
-[100 - x - y] = -20
x + y - 100 = -20
x + y - 100 - y + 100 = -20 - y + 100
x = 80 - y
sqrt(x) + sqrt(y) = 10
sqrt(x) + sqrt(y) - sqrt(y) = 10 - sqrt(y)
sqrt(x) = 10 - sqrt(y)
[sqrt(x)]^2 = [10 - sqrt(y)]^2
x = 10^2 + [-sqrt(y)]^2 + 2*10*[-sqrt(y)]
x = 100 + y - 20*sqrt(y)
100 + y - 20*sqrt(y) = x
Since x = 80 - y, then
100 + y - 20*sqrt(y) = 80 - y
100 + y - 20*sqrt(y) - 80 + y = 80 - y - 80 + y
20 + 2y - 20*sqrt(y) = 0
2y - 20*sqrt(y) + 20 = 0
[2y - 20*sqrt(y) + 20]/2 = 0 / 2
y - 10*sqrt(y) + 10 = 0
Let u = sqrt(y), then
y - 10*sqrt(y) + 10 = u^2 - 10u + 10 = 0
u = [-(-10) +/- sqrt([-10]^2 - 4*1*10)] / [2*1]
u = [10 +/- sqrt(100 - 40)] / 2
u = [10 +/- sqrt(60)] / 2
u = [10 +/- sqrt(4*15)] / 2
u = [10 +/- 2*sqrt(15)] / 2
u = 5 +/- sqrt(15)
u^2 = [5 +/- sqrt(15)]^2
u^2 = 5^2 + [+/- sqrt(15)]^2 + 2*5*[+/- sqrt(15)]
u^2 = 25 + 15 +/- 10*sqrt(15)
u^2 = 40 +/- 10*sqrt(15)
Since u = sqrt(y), then
u^2 = [sqrt(y)]^2 = 40 +/- 10*sqrt(15)
y = 40 +/- 10*sqrt(15)
y1 = 40 + 10*sqrt(15) and y2 = 40 - 10*sqrt(15)
y1 > 0 and y2 > 0
Since x = 80 - y, then
x1 = 80 - y1 and x2 = 80 - y2
x1 = 80 - [40 + 10*sqrt(15)] and x2 = 80 - [40 - 10*sqrt(15)]
x1 = 80 - 40 - 10*sqrt(15) and x2 = 80 - 40 + 10*sqrt(15)
x1 = 40 - 10*sqrt(15) and x2 = 40 + 10*sqrt(15)
x1 > 0 and x2 > 0
{ (x1, y1), (x2, y2) } = { (40 - 10*sqrt[15], 40 + 10*sqrt[15]), (40 + 10*sqrt[15], 40 - 10*sqrt[15]) }

[@wes9627]

√x=5+z and √y=5-z; (5+z)(5-z)=10; z^2=15; z=±√15; x=(5±√15)^2 and y=(5∓√15)^2

[@souvik7752]

Use 4ab