Question

solve for completing the square. x squared -8x+12=0. show work please

some have answered it before but i still do not know how to get to the 16.

Answer 1

do you know how to complete a square?

Answer 2

we are working on that right now in class, but my teacher is doing 3 sections in the book a day three days a week, we are not actually being taught anything,

Answer 3

Ok, Completing the square is not only a method to solve quadratics, but in the future it allows you to easily put the equation in vertex for to find the vertex.

It takes advantage of the fact that a -a = 0. So you can add an expression as long as you subtract it too in an equation.

So if you have a piece of a "square"

\[x^2 + ax\] you know that somewhere there could be a
\[(x+b)^2 = x^2+ax + something\]
The basic equation/process is
\[y=x^2 + ax +c\]
\[y=x^2 +ax +\left(\frac{a}{2}\right)^2- \left(\frac{a}{2}\right)^2 +c\]
and sos
\[y=\left(x+\frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2 +c\]
You can FOIL the \[\left(x+\frac{a}{2}\right)^2 \] to see how it works.

Answer 4

So model example with real numbers (I can't do your problem for you I got warned by mods)

\[x^2 -10x +20=0\]
becomes
\[x^2 -10x +5^2 -(5^2) +20 =0\]
becomes
\[(x-5)^2 -(25)+20 =0\]
becomes
\[(x-5)^2 -5 =0\]
\[(x-5)^2 =5\]
\[x-5 =\pm \sqrt{5}\]

\[x=+5\pm \sqrt{5}\]

Answer 5

ok, do not want the answer given to me, i already have the answer, i am trying to understand the problem and how the 16 comes into it. cause i do not understand. and no one seems to be listening to me when i go to tutoring telling them that i do not understand fractions at all. so i am in college and about to flunk out of math because the teacher is moving at warp speed and i am stupid when it comes to fractions

Answer 6

and since i do not understand what i am doing it is even harder to explain to me using a different problem

Answer 7

Ok the 16 came from the fact that
\[\left(\frac{8}{2}\right)^2 = 4^2 = 16\]

In order to complete the square of
\[x^2 -8x\] you needed to add to the equation
$$ \left(\frac{8}{2}\right)^2-\left(\frac{8}{2}\right)^2 $$ since this is zero and you aren't just deciding to change the equation.

They wanted you to add this to the equation so that you would have
\[x^2 +8x +\left(\frac{8}{2}\right)^2 \] in part of the equation that you could then convert into the format of
\[(x-4)^2\]
since

\[(x-4)^2 = x^2 - 8x + 16\]

So you add a "zero" value of +(8/2)^2 and -(8/2)^2 ... use ONLY the positive one to get to a squared term and then fold the negative one into the rest of the equation.

Does that help at all?

Answer 8

and $$\frac{8}{2} = 4$$ so $$4^2=16$$

Answer 9

thank you so much that does help. especially since my teacher is no help when he is asked questions he looks at you like you are stupid, and so have all the tutors at the school when i went to tutoring.

Answer 10

Ok, Try and go through your problem and make sure it makes sense to you, and then try and go through my example to cement it if you feel comfortable.
I have to go now though. Good luck!