# How do you find the roots, real and imaginary, of y=-x^2+5x-9  using the quadratic formula?

##### 1 Answer
Mar 16, 2016

real roots: none
imaginary roots: $x = \frac{5 \pm \sqrt{11} i}{2}$

#### Explanation:

Since the given equation is already in standard form, identify the $\textcolor{b l u e}{a} , \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{b} ,$ and $\textcolor{v i o \le t}{c}$ values. Then plug the values into the quadratic formula to solve for the roots.

$y = \textcolor{b l u e}{- 1} {x}^{2}$ $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 5} x$ $\textcolor{v i o \le t}{- 9}$

$\textcolor{b l u e}{a = - 1} \textcolor{w h i t e}{X X X X X} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{b = 5} \textcolor{w h i t e}{X X X X X} \textcolor{v i o \le t}{c = - 9}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{5}\right) \pm \sqrt{{\left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{5}\right)}^{2} - 4 \left(\textcolor{b l u e}{- 1}\right) \left(\textcolor{v i o \le t}{- 9}\right)}}{2 \left(\textcolor{b l u e}{- 1}\right)}$

$x = \frac{- 5 \pm \sqrt{25 - 36}}{-} 2$

$x = \frac{- 5 \pm \sqrt{- 11}}{-} 2$

$x = \frac{- \left(- 5 + \sqrt{11} i\right)}{2} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{- \left(- 5 - \sqrt{11} i\right)}{2}$

$x = \frac{5 - \sqrt{11} i}{2} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{5 + \sqrt{11} i}{2}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{5 \pm \sqrt{11} i}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the imaginary roots are $x = \frac{5 \pm \sqrt{11} i}{2}$.