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

How do you find the discriminant and how many solutions does 2x^2-4x+1=02x^2-4x+1=0 have?