What are the solutions to the equation (2y-1)^2+(2y+1)^2=(2y)^2+6 ?

The solutions to the equation (2y-1)^2+(2y+1)^2=(2y)^2+6 are y=1,y=-1

Find the zeros of (2y-1)^2+(2y+1)^2=(2y)^2+6

The zeros of (2y-1)^2+(2y+1)^2=(2y)^2+6 are y=1,y=-1

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(2 y - 1)^{2} + (2 y + 1)^{2} = (2 y)^{2} + 6

y = 1, y = - 1

Solution steps

(2 y - 1)^{2} + (2 y + 1)^{2} = (2 y)^{2} + 6

Expand

(2 y - 1)^{2} + (2 y + 1)^{2} : 8 y^{2} + 2

Expand

(2 y)^{2} + 6 : 4 y^{2} + 6

8 y^{2} + 2 = 4 y^{2} + 6

Move

2

to the right side

8 y^{2} = 4 y^{2} + 4

Move

4 y^{2}

to the left side

4 y^{2} = 4

Divide both sides by

4

y^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = 1, y = - 1

1 = 1

- 1 = - 1

y = 1, y = - 1

(- 6 p^{2} q^{3})^{2}

4 x^{2} + 32 x

(\frac{a ^{3}}{b ^{2}})^{4}

x^{3} + 2 x^{2} \geq 4 x + 8

4^{3} \times (- 3)^{3}

What are the solutions to the equation (2y-1)^2+(2y+1)^2=(2y)^2+6 ?
The solutions to the equation (2y-1)^2+(2y+1)^2=(2y)^2+6 are y=1,y=-1
Find the zeros of (2y-1)^2+(2y+1)^2=(2y)^2+6
The zeros of (2y-1)^2+(2y+1)^2=(2y)^2+6 are y=1,y=-1