What are the solutions to the equation 8x^2-4x+1=0 ?

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

Find the zeros of 8x^2-4x+1=0

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

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

x = \frac{1}{4} + i \frac{1}{4}, x = \frac{1}{4} - i \frac{1}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 8 \cdot 1 : 4 i

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 4 i}{2 \cdot 8} : \frac{1}{4} + i \frac{1}{4}

x = \frac{- ( - 4 ) - 4 i}{2 \cdot 8} : \frac{1}{4} - i \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{1}{4}, x = \frac{1}{4} - i \frac{1}{4}

\frac{( \frac{3}{2} ) ^{6}}{( \frac{3}{5} ) ^{6}}

2 (3 x - 5) = 8

15 x - (10 + 5 x) = 20 x - 10

5 x - 2 = 3 x + 40

(- \frac{π}{2})^{2}

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