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Question #00865

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Neha M. · Stefan V.

Feb 12, 2017

$x = - \frac{1}{2} \pm \frac{\sqrt{3} i}{2}$ $x = -1/2 +- (sqrt(3)i)/2$

Explanation:

$x^{2} + x + 1 = 0$ $x^2 + x + 1 = 0$

The discriminant, $D$ $D$ , is

$D = {(coeff. of x)}^{2} - 4 \times {coeff. of x}^{2} \times constant$ $D = ("coeff. of x")^2 - 4 xx "coeff. of x"^2 xx "constant"$

$D = 1^{2} - 4 (1) (1)$ $D = 1^2 - 4 (1) (1)$

$D = 1 - 4$ $D = 1 - 4$

$D = - 3$ $D = -3$

Solving for $x$ $x$ ; since $D$ $D$ is negative, the equation has complex roots.

$x = \frac{- coeff. of x \pm (\sqrt{D})}{2 \cdot {coeff. of x}^{2}}$ $x = (- "coeff. of x" +- (sqrt(D)))/(2* "coeff. of x"^2)$

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$ $x = (-1 +- sqrt(-3))/(2*1)$

$x = \frac{- 1 \pm \sqrt{3} i}{2}$ $x = (-1 +- sqrt(3)i)/2$

$x = - \frac{1}{2} \pm \frac{\sqrt{3} i}{2}$ $x = -1/2 +- (sqrt(3)i)/2$

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