How do you solve #3x^2 - 12x + 2 = 0#?

1 Answer
May 10, 2017

Answer:

#x = 2+-sqrt(30)/3#

Explanation:

The equation:

#3x^2-12x+2=0#

is in the standard form:

#ax^2+bx+c=0#

with #a=3#, #b=-12# and #c=2#

The discriminant #Delta# of a quadratic is given by the formula:

#Delta = b^2-4ac = (color(blue)(-12))^2-4(color(blue)(3))(color(blue)(2)) = 144-24 = 120 = 2^2*30#

Since #Delta > 0#, the given quadratic equation has two Real roots, but since #Delta# is not a perfect square those roots are irrational.

We can find the roots using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (12+-sqrt(120))/6#

#color(white)(x) = (12+-sqrt(2^2*30))/6#

#color(white)(x) = (12+-2sqrt(30))/6#

#color(white)(x) = 2+-sqrt(30)/3#

That is:

#x = 2+sqrt(30)/3" "# or #" "x = 2-sqrt(30)/3#