What are the solutions to the equation x^2-2x+1/4 =0 ?

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

Find the zeros of x^2-2x+1/4 =0

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

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

x = \frac{2 + 3}{2}, x = \frac{2 - 3}{2}

Decimal Notation

x = 1.86602 \dots, x = 0.13397 \dots

Solution steps

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

Multiply both sides by

4

4 x^{2} - 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 \cdot 1 = 43

x_{1, 2} = \frac{- ( - 8 ) \pm 4 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 3}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 4 3}{2 \cdot 4}

x = \frac{- ( - 8 ) + 4 3}{2 \cdot 4} : \frac{2 + 3}{2}

x = \frac{- ( - 8 ) - 4 3}{2 \cdot 4} : \frac{2 - 3}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 3}{2}, x = \frac{2 - 3}{2}

1 0^{4 l o g_{10} (5)}

\frac{x}{x + 1} < \frac{x}{x - 1}

\frac{8 x - 1}{x} \leq x + 1

216 x^{2} - 512 = 0

3 n^{4} - 147

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