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#?

1 Answer
Jun 3, 2017

Answer:

Roots are #-1+sqrt6# and #-1-sqrt6#

Explanation:

For an equation #ax^2+bx+c=0#, discriminant is #b^2-4ac#

and quadratic formula gives roots as #x=(-b+-sqrt(b^2-4ac))/(2a)#

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

if #b^2-4ac# is a perfect square (i.e. positive as well), roots are rational

if #b^2-4ac# is positive but not a perfect square, roots are irrational and conjugate i.e. of the type #p+-sqrtq#

if #b^2-4ac# is negative, roots are complex conjugate i.e. of the type #p+-iq#

Here we have #2x-5=-x^2# i.e. #x^2+2x-5=0#

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

#b^2-4ac=2^2-4xx1xx(-5)=4+20=24#

and hence roots are irrational and conjugate.

These are #(-2+-sqrt24)/2=-2/2+-(2sqrt6)/2=-1+-sqrt6#