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.

Related questions
See all questions in Solutions Using the Discriminant
Impact of this question
1836 views around the world
You can reuse this answer
Creative Commons License