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

The solutions to the equation (2y)^2=4y are y=1,y=0

Find the zeros of (2y)^2=4y

The zeros of (2y)^2=4y are y=1,y=0

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(2 y)^{2} = 4 y

y = 1, y = 0

Solution steps

(2 y)^{2} = 4 y

Expand

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

4 y^{2} = 4 y

Move

4 y

to the left side

4 y^{2} - 4 y = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 4}, y_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 4}

y = \frac{- ( - 4 ) + 4}{2 \cdot 4} : 1

y = \frac{- ( - 4 ) - 4}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

y = 1, y = 0

a^{2} - 2 a + 2 = 4

4 x^{2} + 5 = 12 x

2 y (4 y + 3) = 9

x^{2} - 9 x + c

180 = 0.05 s^{2} + 1.1 s

What are the solutions to the equation (2y)^2=4y ?
The solutions to the equation (2y)^2=4y are y=1,y=0
Find the zeros of (2y)^2=4y
The zeros of (2y)^2=4y are y=1,y=0