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

The solutions to the equation 8w^2+w+1=0 are w=-1/16+i(sqrt(31))/(16),w=-1/16-i(sqrt(31))/(16)

Find the zeros of 8w^2+w+1=0

The zeros of 8w^2+w+1=0 are w=-1/16+i(sqrt(31))/(16),w=-1/16-i(sqrt(31))/(16)

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8 w^{2} + w + 1 = 0

w = - \frac{1}{16} + i \frac{31}{16}, w = - \frac{1}{16} - i \frac{31}{16}

Solution steps

8 w^{2} + w + 1 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

Simplify

1^{2} - 4 \cdot 8 \cdot 1 : 31 i

w_{1, 2} = \frac{- 1 \pm 31 i}{2 \cdot 8}

Separate the solutions

w_{1} = \frac{- 1 + 31 i}{2 \cdot 8}, w_{2} = \frac{- 1 - 31 i}{2 \cdot 8}

w = \frac{- 1 + 31 i}{2 \cdot 8} : - \frac{1}{16} + i \frac{31}{16}

w = \frac{- 1 - 31 i}{2 \cdot 8} : - \frac{1}{16} - i \frac{31}{16}

The solutions to the quadratic equation are:

w = - \frac{1}{16} + i \frac{31}{16}, w = - \frac{1}{16} - i \frac{31}{16}

z^{3} - 8

0 = 4 + \frac{n}{5}

(- \frac{1}{3}) (\frac{2}{5})

5 \leq 1 - x \leq 9

x^{2} < 4 x

What are the solutions to the equation 8w^2+w+1=0 ?
The solutions to the equation 8w^2+w+1=0 are w=-1/16+i(sqrt(31))/(16),w=-1/16-i(sqrt(31))/(16)
Find the zeros of 8w^2+w+1=0
The zeros of 8w^2+w+1=0 are w=-1/16+i(sqrt(31))/(16),w=-1/16-i(sqrt(31))/(16)