Janu16
  • Janu16
Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
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?
Janu16
  • Janu16
8? I think
Janu16
  • Janu16
probably not

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x^2\pm\color{red}{\rm 2a}x+\color{red}{\rm a}^2 }\)
SolomonZelman
  • SolomonZelman
Does this look familiar to anything?
SolomonZelman
  • SolomonZelman
`(x - a)² = x² - 2a + a²` `(x + a)² = x² + 2a + a²`
Janu16
  • Janu16
umm no. I learned this but can't get the answer
Janu16
  • Janu16
wait i know that
SolomonZelman
  • SolomonZelman
you know those two in gray, right?
Janu16
  • Janu16
sum and difference of a cube??
SolomonZelman
  • SolomonZelman
No cubes: Just how to expand (x+a)^2 and (x-a)^2
Janu16
  • Janu16
ohok
SolomonZelman
  • SolomonZelman
So, we are going to make x^2-8x (by adding some number) into x^2-2a+a^2 form
Janu16
  • Janu16
ok
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\)
SolomonZelman
  • SolomonZelman
So if 2a is 8, than a is 4. Right?
Janu16
  • Janu16
ya
Janu16
  • Janu16
than you square it?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+\color{red}{\rm 4}^2 =(x-\color{red}{\rm 8})^2 }\) \(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+16 =(x-\color{red}{\rm 8})^2 }\) that would be a perfect square trinomial. So we add 16 to the x^2-8x, to make our polynomial a perfect square.
Janu16
  • Janu16
what is this formula called? cause I didn't learn this?
SolomonZelman
  • SolomonZelman
which formula?
Janu16
  • Janu16
the one you are using
SolomonZelman
  • SolomonZelman
oh, don't worry, you can just google (a+b)^2 and all of that will come up
SolomonZelman
  • SolomonZelman
all of those like (a+b)^2 and (a-b)^2 ... it will all show up.
Janu16
  • Janu16
okk
SolomonZelman
  • SolomonZelman
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.
Janu16
  • Janu16
so it would be -8-13+3
SolomonZelman
  • SolomonZelman
you mixed up something a little, how did you get that result/
SolomonZelman
  • SolomonZelman
??
Janu16
  • Janu16
from that equation
Janu16
  • Janu16
wait x- 8 - 13?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\) 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: \(\large\color{black}{ \displaystyle (x-\color{red}{ 4})^2-13 }\)
SolomonZelman
  • SolomonZelman
read it carefully, and if you have questions, then ask.
Janu16
  • Janu16
ohok tysm!! so you just simplied 8 to 4?
SolomonZelman
  • SolomonZelman
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.
Janu16
  • Janu16
ohhhok f(x) = (x − 4)2 − 13 this would be the answer right?
SolomonZelman
  • SolomonZelman
(x-4)\(\color{red}{^2}\)-13 :D
Janu16
  • Janu16
tysm! can you hep with other one on new thread plz?

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