How do you solve $3x (x - 2) = 1$ ?

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Answer 1 · 2015-08-19T12:34:47

The solutions are
$color(blue)(x=(3+2sqrt3)/3$
$color(blue)(x=(3-2sqrt3)/3$

$3x(x−2)=1$

$3x^2−6x=1$

$3x^2−6x-1=0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=3, b=-6, c=-1$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (-6)^2-(4*3*(-1))$

$= 36+12=48$

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

$x = (-(-6)+-sqrt(48))/(2*3) = (6+-4sqrt(3))/6$

$( cancel2(3+-2sqrt(3)))/cancel6 = (3+-2sqrt3)/3$

The solutions are
$color(blue)(x=(3+2sqrt3)/3$
$color(blue)(x=(3-2sqrt3)/3$

How do you solve 3x (x - 2) = 13x (x - 2) = 1?