What is the answer to 2x^2-4x=2x*x-2x^{0.2} ?

The answer to 2x^2-4x=2x*x-2x^{0.2} is x=0,x=-1/(2\sqrt[4]{2)},x= 1/(2\sqrt[4]{2)}

Solve for x: 2x^2-4x=2x*x-2x^{0.2}

The solution is x=0,x=-1/(2\sqrt[4]{2)},x= 1/(2\sqrt[4]{2)}

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2 x^{2} - 4 x = 2 x \cdot x - 2 x^{0.2}

Decimal Notation

x = 0, x = - 0.42044 \dots, x = 0.42044 \dots

Solution steps

2 x^{2} - 4 x = 2 xx - 2 x^{0.2}

Expand

2 xx - 2 x^{0.2} : 2 x^{2} - 2 x^{0.2}

2 x^{2} - 4 x = 2 x^{2} - 2 x^{0.2}

Remove roots

- 1024 x^{5} = - 32 x

Solve

Verify Solutions:

x = 0

True

The solutions are

8 x + 33 - 2 x + 24 = 3

26 = - 1 + (27 x)^{\frac{3}{4}}

4 x - 4 x + 1 = 0

x = 6 + x

- x - 1 = x + 1

What is the answer to 2x^2-4x=2x*x-2x^{0.2} ?
The answer to 2x^2-4x=2x*x-2x^{0.2} is x=0,x=-1/(2\sqrt[4]{2)},x= 1/(2\sqrt[4]{2)}
Solve for x: 2x^2-4x=2x*x-2x^{0.2}
The solution is x=0,x=-1/(2\sqrt[4]{2)},x= 1/(2\sqrt[4]{2)}