What are the solutions to the equation 16x^2-16x+8=0 ?

The solutions to the equation 16x^2-16x+8=0 are x= 1/2+i 1/2 ,x= 1/2-i 1/2

Find the zeros of 16x^2-16x+8=0

The zeros of 16x^2-16x+8=0 are x= 1/2+i 1/2 ,x= 1/2-i 1/2

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16 x^{2} - 16 x + 8 = 0

x = \frac{1}{2} + i \frac{1}{2}, x = \frac{1}{2} - i \frac{1}{2}

Solution steps

16 x^{2} - 16 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

Simplify

(- 16)^{2} - 4 \cdot 16 \cdot 8 : 16 i

x_{1, 2} = \frac{- ( - 16 ) \pm 16 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 16 i}{2 \cdot 16}, x_{2} = \frac{- ( - 16 ) - 16 i}{2 \cdot 16}

x = \frac{- ( - 16 ) + 16 i}{2 \cdot 16} : \frac{1}{2} + i \frac{1}{2}

x = \frac{- ( - 16 ) - 16 i}{2 \cdot 16} : \frac{1}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{1}{2}, x = \frac{1}{2} - i \frac{1}{2}

\frac{9}{27} = \frac{x}{18}

- 5 - (- 9)

(\frac{( t ^{- 5} )}{( t ^{2} t ^{5} )})^{- 4}

(x + 3)^{2} - 4

\frac{7}{3} x = 8

What are the solutions to the equation 16x^2-16x+8=0 ?
The solutions to the equation 16x^2-16x+8=0 are x= 1/2+i 1/2 ,x= 1/2-i 1/2
Find the zeros of 16x^2-16x+8=0
The zeros of 16x^2-16x+8=0 are x= 1/2+i 1/2 ,x= 1/2-i 1/2