A Nice Algebra Problem | Olympiad Mathematics (x,y)=?

[@is7728]

Let a and b be √x and √y respectively.
a + b = 10 and ab = 10
Substituting a = 10 - b into the 2nd eqn,
(10 - b) * b = 10
-b^2 + 10b - 10 = 0
b^2 - 10b + 10 = 0
By the quadratic formula,
b = (10 +- √(100 - 40)) / 2
= 5 +- √15
So for the 1st eqn, a + (5 +- √15) = 10
Then, a + 5 +- √15 = 10
a +- √15 = 5
a = 5 - + √15 while b = 5 + - √15
So far we get 2 sets of solutions, and now recall that a = √x and b = √y
x = (5 - + √15)^2 and y = (5 + - √15)^2
x = 40 - + 10√15 and y = 40 + - 10√15

[@MyHoangthi-v4n]

Creative world design

[@RyanLewis-Johnson-wq6xs]

Sqrt[X]+Sqrt[Y]=10 Sqrt[XY]=10 (X,Y)=(40±10Sqrt[15],40±10Sqrt[15])

[@Segalmed]

Sqrt(x) + SQRT(y) = 10 => (Sqrt(x) + SQRT(y))² = 100 => x + y + 2*SQRT(x)*SQRT(y) = 100 => (SQRT(x)*SQRT(y) = 10 => x + y + 20 = 100 => x + y = 80
x = 80 - y y = 80 - x
SQRT(x*y) = 10 => x*y = 100
Insert x or y as in 2nd line
y² - 80y + 100 = 0 x² - 80x +100 = 0
=> x1,2 = 40 +/- SQRT (1500) y1,2 = 40 +/- SQRT(1500)
Neither both plus nor both minus solutions are valid for the problem but one plus one minus works.
=> x1= 40 + SQRT (1500) y1 = 40 - SQRT(1500) x2= 40 - SQRT (1500) y2 = 40 + SQRT(1500)

[@ZawLinnThein-r6o]

❤ Thank s

[@imetroangola17]

Sendo x+y=S e xy=P, onde S e P são a soma e o produto das raízes de uma equação do 2° grau, respectivamente, cuja equação é representada por:
k²-Sk+P=0. No caso, S=80 e P=100.
∆=80²-4×100=6400-400=6000
√∆=20√15, daí
k=(80±20√15)/2=40 ± 10√15
Portanto, as raízes são:
40+10√15 e 40 - 10√15
Conjunto solução:
*{(40+10√15, 40 -10√15), (40 -10√15, 40+10√15)}*