Question

(x-5)^2+(x-y)^2=0 what is the vale of x and y ?
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Answer 1

Answer:

If x and y are real, and (x - 1)^2 + (y - 4)^2 = 0, then what is the value of x^3 + y^3?

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(X-1)^2+(y-2)^2=0

i.e (x-1)^2=-(y-4)^2

If x-1 is not =0 and y-4 is not =0

Then

Left side is +ve and right side is -ve

i.e +ve =-ve

it is not possible

So x-1=0 and y-4=0

Which satiesfied the above eqn

So x=1 and y=4

Value of x^3+y^3=1^3+4^3=1+64=65

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Real Numbers

The Real Numbers had no name before Imaginary Numbers were thought of. They got called "Real" because they were not Imaginary.

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.

Answer 2

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