What are the solutions to the equation (z+1)^2=2z^2 ?

The solutions to the equation (z+1)^2=2z^2 are z=1-sqrt(2),z=1+sqrt(2)

Find the zeros of (z+1)^2=2z^2

The zeros of (z+1)^2=2z^2 are z=1-sqrt(2),z=1+sqrt(2)

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(z + 1)^{2} = 2 z^{2}

z = 1 - 2, z = 1 + 2

Decimal Notation

z = - 0.41421 \dots, z = 2.41421 \dots

Solution steps

(z + 1)^{2} = 2 z^{2}

Expand

(z + 1)^{2} : z^{2} + 2 z + 1

z^{2} + 2 z + 1 = 2 z^{2}

Move

2 z^{2}

to the left side

- z^{2} + 2 z + 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 1 = 22

z_{1, 2} = \frac{- 2 \pm 2 2}{2 ( - 1 )}

Separate the solutions

z_{1} = \frac{- 2 + 2 2}{2 ( - 1 )}, z_{2} = \frac{- 2 - 2 2}{2 ( - 1 )}

z = \frac{- 2 + 2 2}{2 ( - 1 )} : 1 - 2

z = \frac{- 2 - 2 2}{2 ( - 1 )} : 1 + 2

The solutions to the quadratic equation are:

z = 1 - 2, z = 1 + 2

- 6.5 q^{2} + 390 q - 2000 = 0

x, x^{2} - 3 x y + 2 y^{2} = 0

- 5 x^{2} + 525 x - 6300 = 0

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

y (y + 7) = 0

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