What are the solutions to the equation 4x=x(2x-4)^2 ?

The solutions to the equation 4x=x(2x-4)^2 are x=0,x=1,x=3

Find the zeros of 4x=x(2x-4)^2

The zeros of 4x=x(2x-4)^2 are x=0,x=1,x=3

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4 x = x (2 x - 4)^{2}

x = 0, x = 1, x = 3

Solution steps

4 x = x (2 x - 4)^{2}

Move

x (2 x - 4)^{2}

to the left side

4 x - x (2 x - 4)^{2} = 0

Factor

4 x - x (2 x - 4)^{2} : - 4 x (x - 1) (x - 3)

- 4 x (x - 1) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

The solutions are

x = 0, x = 1, x = 3

x^{4} + 13 x + 36 = 0

3 x^{3} + 2 x^{2} = 12 x + 8

x^{3} = 17

x^{3} + 3 x^{2} + 5 x + 15 = 0

x^{4} - 4 x^{2} + 1 = 0

What are the solutions to the equation 4x=x(2x-4)^2 ?
The solutions to the equation 4x=x(2x-4)^2 are x=0,x=1,x=3
Find the zeros of 4x=x(2x-4)^2
The zeros of 4x=x(2x-4)^2 are x=0,x=1,x=3