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

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Answer 1 · 2018-03-20T10:32:02

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

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 Tony B

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