How do you find the discriminant for #2x^2-5x+20=0# and determine the number and type of solutions?

1 Answer
Apr 18, 2017

Answer:

Complex and conjugate. #x= 1.25 + 2.9047i , x=1.25 - 2.9047i#

Explanation:

#2x^2-5x+20 =0 #. Comparing with standard equation #ax^2+bx+c=0# we get #a=2 , b= -5 ,c=20#. Disciminant #D= b^2-4ac = (-5)^2 -4*2*20 = -135 # If #D=0# roots are equal.
If #D>0# The roots are real.
If #D < 0#The roots are complex in nature and conjugate.
Here the discriminant is #<0#.So roots are complex in nature and conjugate.
#x= -b/(2a) +- sqrt(b^2-4ac)/(2a) = -(-5)/(2*2) +- sqrt( (-5)^2-4*2*20)/(2*2) = 1.25 +- 2.9047i #
Solution: #x= 1.25 + 2.9047i , x=1.25 - 2.9047i# [Ans]