What are the solutions to the equation (4-2y)^2+4y^2=16 ?

The solutions to the equation (4-2y)^2+4y^2=16 are y=2,y=0

Find the zeros of (4-2y)^2+4y^2=16

The zeros of (4-2y)^2+4y^2=16 are y=2,y=0

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(4 - 2 y)^{2} + 4 y^{2} = 16

y = 2, y = 0

Solution steps

(4 - 2 y)^{2} + 4 y^{2} = 16

Expand

(4 - 2 y)^{2} + 4 y^{2} : 16 - 16 y + 8 y^{2}

16 - 16 y + 8 y^{2} = 16

Move

16

to the left side

8 y^{2} - 16 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

(- 16)^{2} - 4 \cdot 8 \cdot 0 = 16

y_{1, 2} = \frac{- ( - 16 ) \pm 16}{2 \cdot 8}

Separate the solutions

y_{1} = \frac{- ( - 16 ) + 16}{2 \cdot 8}, y_{2} = \frac{- ( - 16 ) - 16}{2 \cdot 8}

y = \frac{- ( - 16 ) + 16}{2 \cdot 8} : 2

y = \frac{- ( - 16 ) - 16}{2 \cdot 8} : 0

The solutions to the quadratic equation are:

y = 2, y = 0

x^{2} + 15 x + 39 = 0

3 x^{2} + 24 x + 15 = 0

x, x^{2} + y^{2} + z^{2} - 2 y + 2 z - 2 = 0

(2 x - 3)^{2} = 6

(x + 2)^{2} = 3 x + 7

What are the solutions to the equation (4-2y)^2+4y^2=16 ?
The solutions to the equation (4-2y)^2+4y^2=16 are y=2,y=0
Find the zeros of (4-2y)^2+4y^2=16
The zeros of (4-2y)^2+4y^2=16 are y=2,y=0