Question

How do you find the discriminant and how many solutions does `2j^2 - 3j = -1` have?

Answer 1

discriminant `D= b^2 - 4ac`

we have: `2j^2 - 3j + 1 = 0`

here:
`a =2` , `b =-3`, `c = 1`
(coefficients of `j^2` , `j` and the constant term respectively)

finding `D`:
`D= b^2 - 4ac = (-3^2) - (4 xx 2 xx 1)`
`D= 9-8 = 1`

formula for roots :
`j = (-b +- sqrt D) / (2a)`
`j = (3 +- sqrt 1) / (2 xx 2)`
# j = (3 + 1) / 4 = 4/4 = 1 and (3 -1) /4 = 2/4 = 1/2#
`j` has two solutions:
`j = 1` and `j = 1/2`

the roots are real and unequal.