Question

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

Answer 1

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]