How do you find the zeros, real and imaginary, of $y = - 6 x^{2} - 2 x + 2$ $y=-6x^2-2x+2$ using the quadratic formula?

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Answer 1 · 2016-07-02T20:14:48

The polynomial is of order 2 so has two roots. These are real, distinct and given by:

$x = - \frac{1}{6} - \frac{\sqrt{13}}{6}$ $x = -1/6 - (sqrt(13))/6$ or $x = - \frac{1}{6} + \frac{\sqrt{13}}{6}$ $x = -1/6 + (sqrt(13))/6$

The quadratic formula, for an equation of the form $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ is given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

For this question a = -6, b = -2, c = 2.

$x = \frac{2 \pm \sqrt{4 - 4 (- 6) (2)}}{2 (- 6)}$ $x = (2 +- sqrt(4 - 4(-6)(2)))/(2(-6))$

$x = \frac{2 \pm \sqrt{52}}{- 12} = \frac{2 \pm 2 \sqrt{13}}{- 12}$ $x = (2 +- sqrt(52))/(-12) = (2 +- 2sqrt(13))/(-12)$

$x = - \frac{1}{6} - \frac{\sqrt{13}}{6}$ $x = -1/6 - (sqrt(13))/6$ or $x = - \frac{1}{6} + \frac{\sqrt{13}}{6}$ $x = -1/6 + (sqrt(13))/6$

How do you find the zeros, real and imaginary, of y=−6x2−2x+2 y=-6x^2-2x+2 using the quadratic formula?