Let √x=a, √y=b , a,b>0 a+b=10 ab=10 then....

Once you got x + y = 80，xy = 100， then x and y are two roots of the following eq. z^2 - 80z + 100 = 0 z = (80 ± √(6400 - 400)) / 2 z = (80 ± √6000) / 2 z = (80 ± 20√15) / 2 z = 40 ± 10√15 So，x = 40 + 10√15，y = 40 - 10√15， or x = 40 - 10√15，y = 40 + 10√15.

Very good.

i ❤ Mathematics

4:58 in my country it is considered an error if the fraction sign is not written at the level of the equal sign

👏👏👌👌en ese caso cuándo y=+ , x=-, e inversa. EXCELENTE

Nosebleed... Wala Sa School Throwback...

You seem amazing if you solving it by complex analysis method, can you solving it sir?

Are x and y only real numbers?

No se puede hacer así. Se utiliza una de las ecuaciones dos veces. En el punto 2, utiliza la primera y la segunda operación

In my world letters can never equal numbers.

Simple solution: Substitute √x=5+z and √y=5-z into the second equation and rearrange to z^2=15, which has roots z=±√15 Thus x=(5+z)^2=(5±√15)^2=40±10√15 and y=(5-z)^2=(5∓√15)^2=40∓10√15 Check: √x√y=(5±√15)(5∓√15)=25-15=10

banale

Not very smart to use x as a variable and a multiplier.

@ludmilak9396

28 күн бұрын

А Вы исследуйте второе исходное равенство -- корень из произведения ху! Произведение не равно 0, значит ни один из сомножителей не равен 0, деление на любое из х и у -- допустимая операция.

@RUBENCORCOS

20 күн бұрын

X2 - 10X + 10 = 0