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

The solutions to the equation 2y^2=4-y^2 are y= 2/(sqrt(3)),y=-2/(sqrt(3))

Find the zeros of 2y^2=4-y^2

The zeros of 2y^2=4-y^2 are y= 2/(sqrt(3)),y=-2/(sqrt(3))

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2 y^{2} = 4 - y^{2}

y = \frac{2}{3}, y = - \frac{2}{3}

Solution steps

2 y^{2} = 4 - y^{2}

Move

y^{2}

to the left side

3 y^{2} = 4

Divide both sides by

3

y^{2} = \frac{4}{3}

For

x^{2} = f (a)

the solutions are

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

y = \frac{4}{3}, y = - \frac{4}{3}

\frac{4}{3} = \frac{2}{3}

- \frac{4}{3} = - \frac{2}{3}

y = \frac{2}{3}, y = - \frac{2}{3}

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

5 x - 7 \leq 7 x - 11 \leq 5 x + 9

0 = - 16 x^{2} + 20 x

x + 8 + ∣ 4 - x ∣ = 0

(x + 3)^{2} - 3 = 0

What are the solutions to the equation 2y^2=4-y^2 ?
The solutions to the equation 2y^2=4-y^2 are y= 2/(sqrt(3)),y=-2/(sqrt(3))
Find the zeros of 2y^2=4-y^2
The zeros of 2y^2=4-y^2 are y= 2/(sqrt(3)),y=-2/(sqrt(3))