# How do you find the roots, real and imaginary, of y= x^2 - 6x - (6x+1)^2  using the quadratic formula?

Feb 22, 2017

We have real zeros $\frac{8}{35} + \frac{\sqrt{46}}{35}$ and $\frac{8}{35} - \frac{\sqrt{46}}{35}$

#### Explanation:

The roots of a quadratic function $y = f \left(x\right) = a {x}^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here we have $y = f \left(x\right) = {x}^{2} - 6 x - {\left(6 x + 1\right)}^{2}$

= ${x}^{2} - 6 x - \left(36 {x}^{2} + 12 x + 1\right)$

= ${x}^{2} - 6 x - 36 {x}^{2} - 12 x - 1$

= $- 35 {x}^{2} - 18 x - 1$

Hence its zeros are $\frac{- \left(- 18\right) \pm \sqrt{{\left(- 18\right)}^{2} - 4 \times \left(- 35\right) \times \left(- 1\right)}}{2 \times \left(- 35\right)}$

= $\frac{18 \pm \sqrt{324 - 140}}{- 70}$

= $\frac{18 \pm \sqrt{184}}{- 70}$

= $\frac{18 \pm 2 \sqrt{46}}{- 70}$

i.e $\frac{8}{35} \pm \frac{\sqrt{46}}{35}$

i.e. $\frac{8}{35} + \frac{\sqrt{46}}{35}$ and $\frac{8}{35} - \frac{\sqrt{46}}{35}$