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

The solutions to the equation x^2-4x+4=2x are x=3+sqrt(5),x=3-sqrt(5)

Find the zeros of x^2-4x+4=2x

The zeros of x^2-4x+4=2x are x=3+sqrt(5),x=3-sqrt(5)

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

x = 3 + 5, x = 3 - 5

Decimal Notation

x = 5.23606 \dots, x = 0.76393 \dots

Solution steps

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

Move

2 x

to the left side

x^{2} - 6 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 4 = 25

x_{1, 2} = \frac{- ( - 6 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 5}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 2 5}{2 \cdot 1}

x = \frac{- ( - 6 ) + 2 5}{2 \cdot 1} : 3 + 5

x = \frac{- ( - 6 ) - 2 5}{2 \cdot 1} : 3 - 5

The solutions to the quadratic equation are:

x = 3 + 5, x = 3 - 5

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

18 - 6 x^{2} = 0

x, z = 1 - x^{2} - y^{2}

(x + 4)^{2} + 38 = 47

x^{2} + 196 = 0

What are the solutions to the equation x^2-4x+4=2x ?
The solutions to the equation x^2-4x+4=2x are x=3+sqrt(5),x=3-sqrt(5)
Find the zeros of x^2-4x+4=2x
The zeros of x^2-4x+4=2x are x=3+sqrt(5),x=3-sqrt(5)