How do you find the discriminant and how many solutions does #2x^2-4x+1=0# have?

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Answer 1 · 2018-04-30T07:43:06

#2x^2-4x+1=0#

has two unequal real roots

the discriminant
for

#ax^2+bx+c=0#

is

#Delta=b^2-4ac#

#Delta>0=>" two unequal real roots"#

#Delta=0=>" equal real roots (ie one root)"#

#Delta<0=>" two complex roots"#

#2x^2-4x+1=0#

#a=2, b=-4,c=1#

#Delta=(-4)^2-4xx2xx1#

#Delta=16-8=8#

#:.Delta>0#

we conclude that

#2x^2-4x+1=0#

has two unequal real roots