Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

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Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

Mathematics
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Ok, tell me this: Say, you have: \(x^2-8x\) THEN, what number would you want to add to the end, to make that a perfect square trinomial?
8? I think
probably not

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Other answers:

\(\large\color{black}{ \displaystyle x^2\pm\color{red}{\rm 2a}x+\color{red}{\rm a}^2 }\)
Does this look familiar to anything?
`(x - a)² = x² - 2a + a²` `(x + a)² = x² + 2a + a²`
umm no. I learned this but can't get the answer
wait i know that
you know those two in gray, right?
sum and difference of a cube??
No cubes: Just how to expand (x+a)^2 and (x-a)^2
ohok
So, we are going to make x^2-8x (by adding some number) into x^2-2a+a^2 form
ok
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\)
So if 2a is 8, than a is 4. Right?
ya
than you square it?
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+\color{red}{\rm 4}^2 =(x-\color{red}{\rm 8})^2 }\) \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+16 =(x-\color{red}{\rm 8})^2 }\) that would be a perfect square trinomial. So we add 16 to the x^2-8x, to make our polynomial a perfect square.
what is this formula called? cause I didn't learn this?
which formula?
the one you are using
oh, don't worry, you can just google (a+b)^2 and all of that will come up
all of those like (a+b)^2 and (a-b)^2 ... it will all show up.
okk
We will add 16, but we can NOT just add 16, because we will change the value of the trinomial. So this is what we will do: \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }\) \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }\) \(\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }\) in parenthesis, it is a perfect square trinomial and -16+3 is -13.
so it would be -8-13+3
you mixed up something a little, how did you get that result/
??
from that equation
wait x- 8 - 13?
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\) And we want to get our \(x62-8x\) into that form, by adding a value. We will add 16, but we can NOT just add 16, because we will change the value of the trinomial. So this is what we will do: \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }\) \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }\) \(\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }\) I added a magic zero, so to speak. +16-16, is all I did:) in parenthesis, it is a perfect square trinomial and -16+3 is -13. \(\large\color{black}{ \displaystyle (x^2{~}-{~} 2\times\color{red}{ 4}x{~}+{~}\color{red}{ 4}^2)-13 }\) do you see that x\(^2\)-2a+a\(^2\) form in the parenthesis? We know that `a²-2a+a²=(x-a)²`, and thus we get the following: \(\large\color{black}{ \displaystyle (x-\color{red}{ 4})^2-13 }\)
read it carefully, and if you have questions, then ask.
ohok tysm!! so you just simplied 8 to 4?
oh, the result I showed at the end can't be simplified. You can re-write it to what it was in the beginning if you do the inverse operations.
ohhhok f(x) = (x − 4)2 − 13 this would be the answer right?
(x-4)\(\color{red}{^2}\)-13 :D
tysm! can you hep with other one on new thread plz?

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