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

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

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

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

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

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

Decimal Notation

x = 0.85355 \dots, x = 0.14644 \dots

Solution steps

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}

3 + 5 (x + 4) = 4 (x - 2) + 3

\frac{4 a ^{8}}{2 a ^{4}}

\frac{3 2}{2}

∣ 3 x + 2 ∣ = 4 x + 5

d^{2} - 14 d + 48

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