What are the solutions to the equation 1=x^2-4x+4 ?

The solutions to the equation 1=x^2-4x+4 are x=3,x=1

Find the zeros of 1=x^2-4x+4

The zeros of 1=x^2-4x+4 are x=3,x=1

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

x = 3, x = 1

Solution steps

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

Switch sides

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

Move

1

to the left side

x^{2} - 4 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

x = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

(\frac{1}{3})^{4} (- 3)^{2}

(x + 5) \times 4% = 20

27 x^{3} - 27

- (\frac{9}{4})^{\frac{3}{2}}

4 - \frac{1}{6} x = \frac{5}{2}

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