Question

Factoring quadratic.

Answer 1

Step 5 is what I'm having trouble understanding.

Answer 2

@whpalmer4 , @dmezzullo , @Luigi0210

Answer 3

@myininaya ,

Answer 4

Well you know how they got \(3^2\) right?

Answer 5

lol compassion did you got lost on the website? Or is the site messed up?

Answer 6

^I noticed that too.

Answer 7

@myininaya - the site is messed up today. Apparantly, today is everything is Mathematics day. All my subjects are labeled, "mathematics." So I don't know where I'm at at this point.

@Luigi0210 , yes, but the terms to the left confuse me.

Answer 8

so you know whatever you add one side you add to the other right?

Answer 9

Today is mathematics day.

Yes, Myininaya.

Answer 10

\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

Answer 11

Weird. It is showing I'm in biology.

Answer 12

Where did rx come from?

Answer 13

It is in biology, does the URL also say mathematics? because that would be strange...
What happens when you go to openstudy.com/study#/groups/mathematics

Answer 14

Everything just says, "Mathematics." URLs - even my questions are a mirror of the mathematics group. So if you see me posting in the wrong section, you know why.

Answer 15

My chats are normal, however.

Answer 16

That formula I wrote above is good formula to know for completing the square

Answer 17

What is r?

Answer 18

Also, is there a more formal method that doesn't involve formulas?

Answer 19

I didn't really call anything r.
I called something k.
It is whatever is front of x
like this:
\[x^2+6x+(\frac{6}{2})^2=(x+\frac{6}{2})^2 \]

Answer 20

Isn't 4 in front of x, and not 6?

Answer 21

here is another example:
\[x^2+10x+(\frac{10}{2})^2=(x+\frac{10}{2})^2 \]

Well in that example above 6 is in front of x

Answer 22

now looking at your problem you have
\[x^2-4x+y^2-2y=4\]
\[(x^2-4x+ ?)+(y^2-2y+?)=4+?+?\]

Answer 23

in that first question we are going to put ?

Answer 24

I understand the formula.

(x^2 - 4x + 4) +(y^2 - 2y + 1) = 4 + (4 + 1)

Answer 25

on then what is it?

Answer 26

Because we take half the term next to x and square it. Right?

Answer 27

right

Answer 28

The next step is what I am stuck at. Would we use that formula?

Answer 29

\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]
\[x^2-kx+(\frac{k}{2})^2=(x-\frac{k}{2})^2 \]

so
you have
\[x^2-4x+(\frac{4}{2})^2=(x-\frac{4}{2})^2 \]
and
\[y^2-2y+(\frac{2}{2})^2=(y-\frac{2}{2})^2 \]

Answer 30

by the way 4/2=2 and 2/2=1

Answer 31

When I add in a number to complete the square I never actually simplify that number
it makes it much easier

Answer 32

Oh, so we solve for each parentehsis, and that will simply to step 5?

Answer 33

x^2 + rx + (r/2)^2 = (x + r/2)^2

x^2 + 4x + (4/2)^2 = (x + 4/2)^2

x^2 + 4x + 4 = x + 4

Okay, so this is what I got. From here it looks like I can factor the left side, right?

Answer 34

\[x^2+4x+4=(x+2)^2\]

Answer 35

I get a perfect square.

(x + 2)^2 = x + 4

Answer 36

Woah, wait, where did that x^2 + 4x + 4 come from?

Answer 37

:( no no (x+2)^2 doesn't equal x+4

Answer 38

that's what you wrote i thought

Answer 39

Oh, I see what I did. I didn't square all the terms.

(x - 4/2)^2 = x^2 + 4, right?

Answer 40

no no (x-4/2)^2=(x-2)^2=(x-2)(x-2)
which equals x^2-2x-2x+4
which equals x^2-4x-4

Answer 41

Oh, I thought we squared all the terms inside the parenthesizes first.

Answer 42

nope nope
here is another example
say we start with this equation:
\[x^2+y^2-9x+5y=\frac{-1}{4}\]
and we want to find the center and radius

Answer 43

First step put your terms with x's together
and put your terms with y's together

Answer 44

Oh, it just clicked! I see.

Answer 45

\[x^2-9x+y^2+5y=\frac{-1}{4}\]

Answer 46

Go on..

Answer 47

now for completing the square

Answer 48

\[(x^2-9x+(\frac{9}{2})^2)+(y^2+5y+(\frac{5}{2})^2)=\frac{-1}{4}+(\frac{9}{2})^2+(\frac{5}{2})^2\]
Now this it to complete the square
I do the same thing all the times to complete the square when we have coefficient 1 for x^2

Answer 49

now this allows us to write this as
\[(x-\frac{9}{2})^2+(y+\frac{5}{2})^2=\frac{-1}{4}+\frac{81}{4}+\frac{25}{4} \]

Answer 50

\[(x-\frac{9}{2})^2+(y+\frac{5}{2})^2=\frac{105}{4}\]

Answer 51

you try this one:
\[x^2+y^2+6y=-4\]

Answer 52

Okay.

\[x^2 + (y^2 + 6y) = -4\]\[x^2 + (y^2 + 6y + \left( \frac{ 6 }{ 2 } \right)^2) = (-4 + \left( \frac{ 6 }{ 2 } \right)^2)\]

Answer 53

\[x^2+(y^2+6y+(\frac{6}{2})^2)=-4+(\frac{6}{2})^2 \]
ok that is right good so far

Answer 54

\[x^2+(y+?)^2=-4+3^2\]
\[x^2+(y+?)^2=-4+9\]
\[x^2+(y+?)^2=5\]
what is that ? number

Answer 55

You know what is weird?
the file you attached showing you were in math displayed the biology icon.

Answer 56

Okay, so our next step would be to simplfy?

x^2 + (y^2 + 6y + 9) = -4 + 9

Factor? But this polynomial isn't factorable.

Answer 57

it is
we are doing the same thing everything
we are writting
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

Answer 58

no do write that (6/2)^2 as 9
do not take care of the square part
just write (6/2)^2 as 3^2

so
\[x^2+(y^2+6y+3^2)=-4+9\]

Answer 59

\[x^2+(y+3)^2=5 \]

Answer 60

Oh, okay. But that doesn't change the value?

Answer 61

it doesn't
it is just easier to factor