What is the answer to (1-1)^2+(2y-4)^2=0 ?

The answer to (1-1)^2+(2y-4)^2=0 is y=2

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

(1 - 1)^{2} + (2 y - 4)^{2} = 0

y = 2

Solution steps

(1 - 1)^{2} + (2 y - 4)^{2} = 0

Factor

(1 - 1)^{2} + (2 y - 4)^{2} : 4 (y - 2)^{2}

4 (y - 2)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 2 = 0

Solve

y - 2 = 0 : y = 2

The solution is

y = 2

(x + 4) \cdot (y - 2) = 0

2 e w s^{2} = 0

x^{2} + 2 x y p a r a x = 0

3 x^{2} y^{4} = 0, 4 x^{3} y^{3} = 0

2 ab = 0

What is the answer to (1-1)^2+(2y-4)^2=0 ?
The answer to (1-1)^2+(2y-4)^2=0 is y=2
Solve for y: (1-1)^2+(2y-4)^2=0
The solution is y=2