How do you find the roots of #x^2-6x+12=0#?

1 Answer
Mar 20, 2018

Answer:

#x=3+-sqrt(3)color(white)(..)i#

Explanation:

Given: #0=x^2-6x+12#

#color(brown)("The question specifically equates to 0. So we must find a solution that")##color(brown)("works for "y=0. #

Consider #y=ax^2+bx+c color(white)("ddd")and color(white)("ddd")x=(-b+-sqrt(b^2-4ac))/(2a)#

Now lets look at the equation part : #b^2-4ac#. THis is called the determinant.

#b^2-4ac color(white)("ddd")->color(white)("d") color(white)("ddd") (-6)^2-4(1)(12) = -12#

As this is negative the graph does not have any x-intercepts where #x# is in the set of what is called real numbers #x !in RR#

There will be a solution of #x# where it is in the set of 'Complex Numbers. #x in CC#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x=(-b+-sqrt(b^2-4ac))/(2a) color(white)("d")->color(white)("d")x=(+6+-sqrt(-12))/2#

#x=3+-sqrt((-12)/4)#

#x=3+-sqrt(3xx(-1))#

#x=3+-sqrt(3)xxsqrt(-1)#

But #sqrt(-1)=i#

#x=3+-sqrt(3)color(white)(..)i#

Tony B