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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.
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
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.
Let √x=a, √y=b , a,b>0
a+b=10
ab=10 then....
i ❤ Mathematics
Very good.
4:58 in my country it is considered an error if the fraction sign is not written at the level of the equal sign
Nosebleed... Wala Sa School Throwback...
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.
👏👏👌👌en ese caso cuándo y=+ , x=-, e inversa. EXCELENTE
You seem amazing if you solving it by complex analysis method, can you solving it sir?
No se puede hacer así. Se utiliza una de las ecuaciones dos veces. En el punto 2, utiliza la primera y la segunda operación
Are x and y only real numbers?
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.
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.
X2 - 10X + 10 = 0