How do you solve $2x^2 - 4x + 1 = 0$ ?

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Answer 1 · 2015-08-21T07:36:49

The solutions are:
$color(blue)(x=(2+sqrt(2))/2$
$color(blue)(x=(2-sqrt(2))/2$

$2x^2-4x+1$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=2, b=-5, c=1$

The Discriminant is given by:
$Delta=b^2-4*a*c$
$= (-4)^2-(4*2*1)$
$= 16-8=8$

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

$x = (-(-4)+-sqrt(8))/(2*2) = (4+-2sqrt(2))/4$

$=(2 (2+-sqrt(2)))/4 = (2+-sqrt(2))/2$

The solutions are:

$color(blue)(x=(2+sqrt(2))/2$
$color(blue)(x=(2-sqrt(2))/2$

How do you solve 2x^2 - 4x + 1 = 02x^2 - 4x + 1 = 0?