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

1 Answer
Binayaka C.
Apr 18, 2017

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 < 0The 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]

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