What are the solutions to the equation 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 ?

The solutions to the equation 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 are x=sqrt(2)-1,x=0

Find the zeros of 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2

The zeros of 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 are x=sqrt(2)-1,x=0

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

x = 2 - 1, x = 0

Solution steps

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

Expand

2 (- x + 2 x) : - 2 x + 22 x

Expand

(x - (x + 2 x))^{2} : 2 x^{2}

- 2 x + 22 x = 2 x^{2}

Switch sides

2 x^{2} = - 2 x + 22 x

Subtract

- 2 x + 22 x

from both sides

2 x^{2} - (- 2 x + 22 x) = - 2 x + 22 x - (- 2 x + 22 x)

Simplify

2 x^{2} + (2 - 22) x = 0

Solve with the quadratic formula

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

(2 - 22)^{2} - 4 \cdot 2 \cdot 0 = 22 - 2

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 - 1, x = 0

2 x (x - 8) = (x + 1) (2 x - 3)

(2 x)^{4}

(0.6)^{2}

\frac{3 + 2 i}{6 + 5 i}

What are the solutions to the equation 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 ?
The solutions to the equation 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 are x=sqrt(2)-1,x=0
Find the zeros of 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2
The zeros of 2(-x+sqrt(2)x)=(x-(x+sqrt(2)x))^2 are x=sqrt(2)-1,x=0