What are the solutions to the equation (0+1)^2+(((y-1)^2)/2)=1 ?

The solutions to the equation (0+1)^2+(((y-1)^2)/2)=1 are y=1

Find the zeros of (0+1)^2+(((y-1)^2)/2)=1

The zeros of (0+1)^2+(((y-1)^2)/2)=1 are y=1

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

y = 1

Solution steps

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

Move

(0 + 1)^{2}

to the right side

\frac{( y - 1 ) ^{2}}{2} = 0

Multiply both sides by

2

(y - 1)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

y - 1 = 0 : y = 1

The solution to the quadratic equation is:

y = 1

x^{2} - 2 x + 96 = 0

12 k^{2} - 32 k - 64 = 0

(x) = x^{2} - 6 x + 5

x = 3 x^{2} - 2, y = \frac{ln ( x ) , d y}{d x}

- 5 x^{2} + 12 x - 4 = 0

What are the solutions to the equation (0+1)^2+(((y-1)^2)/2)=1 ?
The solutions to the equation (0+1)^2+(((y-1)^2)/2)=1 are y=1
Find the zeros of (0+1)^2+(((y-1)^2)/2)=1
The zeros of (0+1)^2+(((y-1)^2)/2)=1 are y=1