# How do you find the roots, real and imaginary, of y= -13x^2 -5x +11(x-3)^2  using the quadratic formula?

Aug 4, 2016

We have two irrational zeros $- \frac{71}{4} - \frac{1}{4} \sqrt{5833}$ and $- \frac{71}{4} + \frac{1}{4} \sqrt{5833}$

#### Explanation:

$y = - 13 {x}^{2} - 5 x + 11 {\left(x - 3\right)}^{2}$

= $- 13 {x}^{2} - 5 x + 11 \left({x}^{2} - 6 x + 9\right)$

= $- 13 {x}^{2} - 5 x + 11 {x}^{2} - 66 x + 99$

= $- 2 {x}^{2} - 71 x + 99$

As discriminant ${b}^{2} - 4 a c = {\left(- 71\right)}^{2} - 4 \times \left(- 2\right) \times 99$

= $5041 + 792 = 5833$ which is positive but not a square of a rational number and hence we have irrational numbers as zeros of te function $y = - 13 {x}^{2} - 5 x + 11 {\left(x - 3\right)}^{2}$ and these are given by quadratic formula

$\frac{- \left(- 71\right) \pm \sqrt{5833}}{- 4} = - \frac{71}{4} \pm \frac{1}{4} \sqrt{5833}$ or

$- \frac{71}{4} - \frac{1}{4} \sqrt{5833}$ and $- \frac{71}{4} + \frac{1}{4} \sqrt{5833}$

Note: Only factors of $5833$ are $19$ and $307$.