Question

(x-2)^2+(x+y)^2=0 then what is the value of y?​

Answer 1

Step-by-step explanation:

Means X and Y can be any number but not imaginary.

(x - 1)^2 + (y - 4)^2 = 0

first solution who satisfy this condition:

(x - 1)^2 =0 , x=1 ; (y - 4)^2=0 ,y=4; x^3 + y^3 = 1+64=65

then finding another solution :

(x - 1)^2 =1,1+-sqrt(1) then must be fitting this condition (y - 4)^2=-1 , it gives imaginary number where for our Y is real number so not satisfy this condition.

So only one solution are there and That is 65