What are the solutions to the equation (-2y)^2+y^2=5 ?

The solutions to the equation (-2y)^2+y^2=5 are y=1,y=-1

Find the zeros of (-2y)^2+y^2=5

The zeros of (-2y)^2+y^2=5 are y=1,y=-1

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(- 2 y)^{2} + y^{2} = 5

y = 1, y = - 1

Solution steps

(- 2 y)^{2} + y^{2} = 5

Expand

(- 2 y)^{2} + y^{2} : 5 y^{2}

5 y^{2} = 5

Move

5

to the left side

5 y^{2} - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

0^{2} - 4 \cdot 5 (- 5) = 10

y_{1, 2} = \frac{- 0 \pm 10}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 0 + 10}{2 \cdot 5}, y_{2} = \frac{- 0 - 10}{2 \cdot 5}

y = \frac{- 0 + 10}{2 \cdot 5} : 1

y = \frac{- 0 - 10}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

y = 1, y = - 1

5 x^{2} + 25 = 0

9 x^{2} + 24 x - 7 = 0

2 x^{2} = 3 x - 10

a, (a^{3} + 27) = (a + 3) (a^{2} + ma + 9)

6 x^{2} - 84 x + 228 = 0

What are the solutions to the equation (-2y)^2+y^2=5 ?
The solutions to the equation (-2y)^2+y^2=5 are y=1,y=-1
Find the zeros of (-2y)^2+y^2=5
The zeros of (-2y)^2+y^2=5 are y=1,y=-1