# How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given 2x-5=-x^2?

Jun 3, 2017

Roots are $- 1 + \sqrt{6}$ and $- 1 - \sqrt{6}$

#### Explanation:

For an equation $a {x}^{2} + b x + c = 0$, discriminant is ${b}^{2} - 4 a c$

and quadratic formula gives roots as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Note that if $a , b$and $c$ are rational number, as we have in the given example

if ${b}^{2} - 4 a c$ is a perfect square (i.e. positive as well), roots are rational

if ${b}^{2} - 4 a c$ is positive but not a perfect square, roots are irrational and conjugate i.e. of the type $p \pm \sqrt{q}$

if ${b}^{2} - 4 a c$ is negative, roots are complex conjugate i.e. of the type $p \pm i q$

Here we have $2 x - 5 = - {x}^{2}$ i.e. ${x}^{2} + 2 x - 5 = 0$

and as $a = 1$, $b = 2$ and $c = - 5$, the discriminant is

${b}^{2} - 4 a c = {2}^{2} - 4 \times 1 \times \left(- 5\right) = 4 + 20 = 24$

and hence roots are irrational and conjugate.

These are $\frac{- 2 \pm \sqrt{24}}{2} = - \frac{2}{2} \pm \frac{2 \sqrt{6}}{2} = - 1 \pm \sqrt{6}$