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

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Answer 1 · 2015-08-21T07:45:03

The solutions are:
$x=color(blue)((1-sqrt3)/2$

$x=color(blue)((1+sqrt3)/2$

$2x^2−2x=1$

$2x^2−2x-1=0$

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

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (-2)^2-(4*2*(-1))$

$= 4+8=12$

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

$x = (-(-2)+-sqrt(12))/(2*2) = (2+-2sqrt(3))/4$

$(2+-2sqrt(3))/4= (2(1+-sqrt3))/4$

$=(cancel2(1+-sqrt3))/cancel4$

$=(1+-sqrt3)/2$

The solutions are:
$x=color(blue)((1-sqrt3)/2$
$x=color(blue)((1+sqrt3)/2$

How do you solve 2x^2 - 2x = 12x^2 - 2x = 1?