How do you solve $x^{2} - 6 x + 6 = 0$ $x^2 - 6x + 6 = 0$ ?

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Answer 1 · 2015-08-02T06:27:47

The solutions are:
$x = 3 + \sqrt{3}$ $color(blue)(x=3+sqrt(3)$
$x = 3 - \sqrt{3}$ $color(blue)(x=3-sqrt(3)$

$x^{2} - 6 x + 6$ $x^2-6x+6$

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

The Discriminant is given by:
$Delta=b^2-4*a*c$
$= (-6)^2-(4*(1)*6)$
$= 36-24 = 12$

As $Delta>0$ there are two solutions.

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

$x = (-(-6)+-sqrt(12))/(2*1) = (6+-sqrt(12))/2$

$sqrt12=sqrt(2*2*3) = 2sqrt3$

$x=(6+-2sqrt(3))/2$
$x=(cancel2(3+-sqrt(3)))/cancel2$
$x=3+-sqrt(3)$

The solutions are:
$color(blue)(x=3+sqrt(3)$
$color(blue)(x=3-sqrt(3)$

How do you solve x^2 - 6x + 6 = 0x2−6x+6=0x^2 - 6x + 6 = 0?