Completing the square

[Chase]

Hello, I was just wondering how to solve

3x^2 -6x +4

I have issues with the 3 in front of the equation.

Additionally, my friend would like an easy understandable summary of how to complete the square. Once upon a time it made sense, but now it's like french vocab, foreign and hard to remember.

THANK YOU

[sweetnsimple786]

There are no real roots for that equation, assuming that you've set it to zero. You can use the quadratic formula to get the imaginary roots, which is [ -b +/- sqrt (b^2 - 4ac) ] / 2

If you're not familiar with it, I'm sure there are some friendly pages on Google that'll explain it perfectly.

To complete the square for 3x^2-6x+4=0,

change the form to 3x^2 - 6x = -4

then factor out a three on the left side so that you have 3(x^2 - 2x + ____ ) =-4

You should always have a one as the coefficient for the x^2 term.

Then you need to take the coefficient of the x term, divide it by two, square this number, and add it to both sides: 3(x^2 - 2x + 1) = -4 + 3

**On the left side, you added 1 inside the parentheses, but you multiplied that 1 by 3 outside the parentheses, which is why you add three to the right side.

Now you simplify what's in the parentheses: 3(x-1)^2=1

[victimofIB]

if its in paper2 just put the eqn into the calc and plot a graph, the x-intercepts are the roots of the eqn.