# How do you find the zeros, real and imaginary, of y= -9x^2-2x-3 using the quadratic formula?

Feb 11, 2017

Zeros are $\frac{1}{9} - i \frac{\sqrt{26}}{9}$ and $\frac{1}{9} + i \frac{\sqrt{26}}{9}$, two complex conjugate numbers

#### Explanation:

Zeros of a quadratic function $y = f \left(x\right) = - 9 {x}^{2} - 2 x - 3$ are those values of $x$ for which $y = 0$.

Using quadratic formula for a function $f \left(x\right) = a {x}^{2} + b x + c$,

zeros are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

and hence for $y = f \left(x\right) = - 9 {x}^{2} - 2 x - 3$,

zeros are $\frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \times \left(- 9\right) \times \left(- 3\right)}}{2 \times \left(- 9\right)}$

i.e. $\frac{2 \pm \sqrt{4 - 108}}{- 18} = \frac{2 \pm \sqrt{- 104}}{18}$

= $\frac{2 \pm \sqrt{{2}^{2} \times 26 \times {i}^{2}}}{18}$ - as ${i}^{2} = - 1$

= $\frac{2 \pm i 2 \sqrt{26}}{18} = \frac{1}{9} \pm i \frac{\sqrt{26}}{9}$

Hence zeros are $\frac{1}{9} - i \frac{\sqrt{26}}{9}$ and $\frac{1}{9} + i \frac{\sqrt{26}}{9}$, two complex conjugate numbers