Question

When solving this equation by completing the square, which of the following shows the equation with the square completed?
x2 - 4x - 32 = 0
A. (x - 4)^2 = 36
B. (x - 2)^2 = 36
C. (x - 2)^2 = 36

Answer 1

coefficient of term "x" determines what to add/subtract to complete the square.

Answer 2

Completing a square requires you to have a square (not surprisingly). So you need the (a-b)^2 form again. However, with the (a-b)^2 rule, you can't have a negative last term. So move the 32 over to the right side of the equation

Answer 3

GT is right, the coeff of x does determine what extra numbers you need to add/subtract for the whole "complete the square" process.

The "formula" to get the last term of the right side of the equation is (|b|/2)^2. B is -4, so putting -4 into the formula yields 4. Make sense?

Answer 4

No

Answer 5

Shoot. Where did I lose you?

Answer 6

All of it Algebra to me is one of the most confusing subjects and it's hard for me to grasp

Answer 7

haha, yeah we all have our strengths and weaknesses. Are you more of an English/literature person then?
Anyway, according to (a-b)^2 = a^2 - 2ab + b^2, and considering that the left side of our equation looks like x^2-4x, wouldn't you say that b
You know what? I'm going to try a new approach. Hold on a sec.

Answer 8

Im more a physical heath and safety with sex ed type of person..

Answer 9

\[x^{2} - 4x - 32 = 0\]\[x^{2} - 4x = 32 \]\[x^{2} - 4x + [blank] = 32 + [blank]\]
To find [blank] use \[({|[coefficient\ of\ x]| \over 2})^{2}\]so [blank] = \[({|-4| \over 2})^{2}=({4 \over 2})^{2} = 2^{2} = 4\]
So putting it all together, \[x^{2} - 4x +4 = 32 +4\]

Answer 10

maybe i should try too lol

y=ax^2 +bx +c

convert it perfect squared form

now (x+b)^2 +C thats the form

where b=1/2 * coefficient of x

C=c-(1/4*coefficient of b ^2 )

Answer 11

And those subjects (phys health & sex safety) are much more interesting than the generic math vs reading. Cool!

And math once more:
You need to factor the x^2 - 4x + 4. To do so (this is the traditional factoring method btw), look for factors of 4 that add up to -4.
Factors of 4: 1, 2, 4 AND -1,-2,-4
Of the above factors, only -2 and another -2 will yield a sum of -4.
So your equation looks like this now:\[(x-2)^2=32+4=36\]

Answer 12

Creative I will finish up with u l8ter I got to go to work

Answer 13

ok, hope you get it later at least :)

Answer 14

yeah

Answer 15

To start with... completing square is actually making a quadratic equation into the form
(x+constant)^2 = a constant.
Let's say, it is (x+z)^2 = y
Get this part?

Answer 16

, where z and y are constants

Answer 17

The way you put it Kind of yes

Answer 18

Now, here comes the problem, how to make it into that form? <- your question actually.
To begin with, you need to know this identity:
\[(a \pm b)^2 = a^2\pm2ab+b^2\]
Got it?

Answer 19

But I am confused where everything go

Answer 20

Oh... first you need to understand that identity... Do you understand that identity ?

Answer 21

No

Answer 22

:|
Okay, let's do it.
\[(a+b)^2 = (a+b)(a+b) = a(a+b) +b(a+b)\]\[ = a^2 + ab+ab+b^2 = a^2 + 2ab+b^2\]
Got it?

Answer 23

Oh yeah

Answer 24

So, can you try expanding \[(a-b)^2\]? Let see what you've got for it :)

Answer 25

ok

Answer 26

@PsYcHoEmOmAtH

Answer 27

@PsYcHoEmOmAtH What have you got so far?

Answer 28

-4-4^8-32=0

Answer 29

Huh?
I was asking you to expand \[(a-b)^2\] ...

Answer 30

As you don't understand the identity, we must do it once first, make sure you understand it, knowing it how it comes

Answer 31

One minute please my friend is trying to help me out here in class

Answer 32

Hmm.. Sorry to leave you alone .. But I have to sleep now.

Answer 33

Its ok bye