# How do you find the zeros of  y = -x^2 - 1/11x + 1/7  using the quadratic formula?

Aug 3, 2016

Zeros are $\frac{7}{22} - \frac{7}{2} \sqrt{\frac{491}{847}}$ and $\frac{7}{22} + \frac{7}{2} \sqrt{\frac{491}{847}}$

#### Explanation:

Quadratic formula gives the zeros of $y = a {x}^{2} + b x + c$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence zeros of $y = - {x}^{2} - \frac{1}{11} x + \frac{1}{7}$ are

$x = \frac{- \left(- \frac{1}{11}\right) \pm \sqrt{{\left(- \frac{1}{11}\right)}^{2} - 4 \left(- 1\right) \left(\frac{1}{7}\right)}}{2 \left(\frac{1}{7}\right)}$

= $\frac{\left(\frac{1}{11}\right) \pm \sqrt{\frac{1}{121} + \frac{4}{7}}}{\frac{2}{7}}$

= $\left(\frac{1}{11} \pm \sqrt{\frac{7 + 484}{847}}\right) \times \frac{7}{2}$

= $\frac{7}{22} \pm \frac{7}{2} \sqrt{\frac{491}{847}}$

Hence zeros are $\frac{7}{22} - \frac{7}{2} \sqrt{\frac{491}{847}}$ and $\frac{7}{22} + \frac{7}{2} \sqrt{\frac{491}{847}}$