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

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Answer 1 · 2015-05-05T07:48:55

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:

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

The Disciminant is given by :
$Delta=b^2-4*a*c$
$= (-2)^2-(4*4*1)$
$= 4-16=-12$

If $Delta=0$ then there is only one solution.
(for $Delta>0$ there are two solutions,
for $Delta<0$ there are no real solutions)

As $Delta = -12$ , this equation has NO REAL SOLUTIONS

Note :
The solutions are normally found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

Substituting the value of $Delta$ will give us 2 imaginary roots/solutions.

How do you find the discriminant and how many and what type of solutions does 4x^2-2x+1=04x^2-2x+1=0 have?