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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