What are the solutions to the equation x^2-x+1/8 =0 ?

The solutions to the equation x^2-x+1/8 =0 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4

Find the zeros of x^2-x+1/8 =0

The zeros of x^2-x+1/8 =0 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4

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x^{2} - x + \frac{1}{8} = 0

x = \frac{2 + 2}{4}, x = \frac{2 - 2}{4}

Decimal Notation

x = 0.85355 \dots, x = 0.14644 \dots

Solution steps

x^{2} - x + \frac{1}{8} = 0

Multiply both sides by

8

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 8 \cdot 1 = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{2 + 2}{4}, x = \frac{2 - 2}{4}

x, 0 = x (1 - x) + m - h x

6 x^{2} - 5 x + 4 = 4 (3 x^{2} - 2 x + 1)

t, h = - 16 t^{2} + v^{0} t + h^{0}

7 x^{2} + 8 x - 3 = 0

- x^{2} + 7 x = 10

What are the solutions to the equation x^2-x+1/8 =0 ?
The solutions to the equation x^2-x+1/8 =0 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4
Find the zeros of x^2-x+1/8 =0
The zeros of x^2-x+1/8 =0 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4