Given #2x^2-1=2x#, how do you find the discriminant and the number of solutions?

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Answer 1 · 2018-03-29T12:04:51

Solution: # x = 1/2+sqrt3/2 and x = 1/2-sqrt3/2 #

#2x^2-1=2x or 2x^2-2x-1=0# Comparing with standard

quadratic equation #ax^2+bx+c=0# we get,

# a=2 ,b=-2 ,c=-1# Discriminant # D= b^2-4ac# or

#D=4+8 =12# If discriminant positive, we get two real

solutions, if it is zero we get just one solution, and if it is negative

we get complex solutions.Discriminant is positive here , so it has

two real roots . Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (2+-sqrt12)/4 = (2+-2sqrt3)/4= 1/2+-sqrt3/2#

Solution: # x = 1/2+sqrt3/2 and x = 1/2-sqrt3/2 # [Ans]