Find all real or complex solutions for 3x^2=6x-12

Please show the work

(this is a homework question for my daughter, so I don't think I am
breaking any Google rules; I am trying to help her)

Hello and thank you for your question.

A first step is to rewrite the equation as equal to zero, that is, adding
(-6x + 12) to both sides:

3x^2 - 6x + 12 = 0

This puts it in the standard form ax^2 + bx + c =0

Now for convenience we'll divide each term by 3
(this is not a necessary step but it will keep the numbers smaller as we
proceed)

x^2 - 2x + 4 = 0
so in terms of the standard form
a=1  b= -2  c= +4


At this point you might spend some time trying to factor it, but I can
assure you that it is not factorable because the "discriminant" b^2 -
4*a*c = 4 - 16 = -12 which is a negative number and hence for the reason
shown below there
are no real number solutions.

So we proceed by substituting these values in the quadratic formula

http://www.cs.colorado.edu/~main/intro/source/quadratic/quadratic.gif

The image above shows the two solutions of the quadratic formula--in text
form they are:
a=1  b= -2  c= +4
x = (-b  +/- sqrt(b^2 - 4*a*c))/2a

x = (2 +/- sqrt(-2^2 - 4 * 1 * 4))/2
x = (2 +/- sqrt(-12))/2

Look again at sqrt(-12)
There is no real number that when squared produces a negative number
Rather sqrt(-12)= i * sqrt(12)  where i = sqrt (-1)
and because 12 = 3*4 we can rewrite sqrt(12) as 2 * sqrt(3)
    (the 4 'inside' the sqrt becomes 2 'outside' the sqrt)

so x = (2 +/- sqrt(-12))/2
becomes x = (2 +/- 2 * i * sqrt(3))/2
or x = 1 +/- i * sqrt(3)

Answer:
The two solutions (both complex) are
x = 1 + i * sqrt(3)
and
x = 1 - i * sqrt(3)

You may find the discussion at the following site useful:
http://www.glencoe.com/sec/math/t_resources/keyconcepts/pdfs/alg103kc_lesson32.p
df

Please excuse the delay in providing this answer to you.

Search terms used
quadratic discriminant "b^2"

Thank you again for bringing us your question.  If you find any of it
unclear, please request clarification.  I would appreciate it if you would
hold off on rating my answer until I have a chance to reply.

Sincerely,
Google Answers Researcher
Richard-ga

---

Clarification of Answer by richard-ga on 01 Oct 2003 11:42 PDT

There's a link in my answer,

http://www.glencoe.com/sec/math/t_resources/keyconcepts/pdfs/alg103kc_lesson32.pdf 

in which the 'df' did not get highlighted as part of an active link.

You may need to add that manually before the link will work properly.

thanks
-R

Richard, perfect answer, thank you. My daughter could fully replicate
your steps.

Many thanks also to grammatoncleric-ga who presented an ingenious 
solution without using the quadratic formula.

One technique that you can use to solve this equation without using
the quadratic formula is to look for factors in the equation.  See
below

3x^2=6x-12

Divide by 3 on both sides:

x^2 = 2x - 4

Subtract (2x - 4) from both sides to equate the left expression to
zero:

x^2 - 2x + 4 = 0

You may see that x^2 - 2x + 1 is easily factored into (x - 1)^2, so
subtract 3 from both sides to get the expression (x^2 - 2x + 1) by
itself:

x^2 - 2x + 1 = -3

Factor the expression (x^2 - 2x + 1):

(x - 1)^2 = -3

Take the square root of both sides, which renders two roots for (x -
1):

(x - 1) = +/- sqrt(-3)

Pull out the factor, i, from the negative square root:

(x - 1) = +/- i*sqrt(3)

Add 1 to both sides:
x = 1 +/- i*sqrt(3)

Both solutions to the equation are complex:

x = 1 + i*sqrt(3), x = 1 - i*sqrtA(3)

The Grammaton Cleric