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

1 Answer
Don't Memorise
May 10, 2015

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.

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