Question

Solve the equation:
11q^2-44=0

Answer 1

what do you get if you divide both sides (all terms) by 11 ?

Answer 2

q^2-4
my teacher taught me how to factor stuff like 16s^2+8s+1 (which is (8s+2)(8s-1)) but I don't know what the =0 part has to do with anything.

Answer 3

you have an equation. Often, you use factoring to solve equations

to solve this equation, I would (as a first step) divide both sides by 11

or if you like, factor an 11 out of each term on the left side

Answer 4

would I have \[11(q^2-4) = 0\] ?

Answer 5

yes. notice you have a difference of squares, so you can factor q^2 - 4
use the rule
\[ (a^2 - b^2) =(a+b)(a-b) \]

Answer 6

quadratic?

Answer 7

I've never seen that formula.
And yes kind of

Answer 8

Because wouldn't factoring 11 out of both sides mean I would have = 0/11

Answer 9

difference of squares arises from (a+b)(a-b) (if you multiply it out, you get a^2 - b^2 )

Answer 10

Oh so you mean how there's no b because it ended up being 0?

Answer 11

**
Because wouldn't factoring 11 out of both sides mean I would have = 0/11
***

we "do the same thing" to both sides of an equation, but that does NOT include factoring. You can think of factoring as re-writing ... it does not change the value ... just how it looks. In other words if you have
2*1+2*2 = 6
you would re-write the left side by factoring out a 2:
2*(1+2) = 6
notice it is still equal. You do not have to also factor the 6

Answer 12

Okay so its basically the same as factoring things without the =0

Answer 13

yes, but in this case, we will actually "solve" the equation for values of q that make the equation true.

so far you have
\[ 11(q^2-4)= 0 \]
now factor q^2 - 4

Answer 14

\[11(q-2)(q+2) = 0\]

Answer 15

here is an example
http://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-special-products/v/factoring-difference-of-squares

Answer 16

do u understand?

Answer 17

now the last part. It is an interesting (but maybe not obvious) idea that if you have
A*B=0
A times B is 0
then either A is zero or B is zero (or maybe both A and B are zero)

we use that idea to solve
\[ 11 (q-2) (q+2) = 0 \]

Answer 18

I've never seen that
and I'll have to go upstairs to watch that video give me a minute

Answer 19

@ilikequadraticsokay

Answer 20

you can watch the video when you have time, it's for future reference.

Answer 21

hello?

Answer 22

@phi
@ilikequadraticsokay

Answer 23

here is the working out and answer

Answer 24

I appreciate that picture but it's not the quadratic formula

Answer 25

Can you just show me how to do it? SOrry if I sound pushy, it's just i'm running out of time:/

Answer 26

I understand that difference of squares, just not how to solve for q

Answer 27

11q^2-44=0 factor out of 11
11(q^2-4)=0 divide both sides by 11.
q^2-4=0
add four to both sides, and you can do it from there.

Answer 28

@ilikequadraticsokay, good?

Answer 29

q^2=4
q=2
??????

Answer 30

yes, but it is ±2.

Answer 31

What is that symbol? I haven't seen it yet.
Do you think y teacher would be okay with me putting =2? She hasn't taught us this material yet, just how to factor....

Answer 32

it means (plus minus)
that basically says that number negative or positive.
In this case it means that the answer is both, negative 2, and positive 2.

Answer 33

Ah, that's in reference to parabolas right?

Answer 34

If your teacher taught you how to factor, I think you should know this.
You can plug in both, it doesn't matter b/c\[(-a)^2=a^2\] and so it is with 2\[(-2)^2=2^2\]

Answer 35

you will get the exact same result if you plug in '2' or '-2'

if the teacher asks you to check, then plug in both and state that each works.

Answer 36

How do i solve for x in (4x + 1)(4 - 1)=0?

Answer 37

@SolomonZelman

Answer 38

yes....

Answer 39

How do i solve for x in (4x + 1)(4 - 1)=0?

Answer 40

if you see that\[a \times b=0\]

it would be true that either a=0 or b=0 @ilikequadraticsokay, right?

Answer 41

I guess?

Answer 42

here you have a similar thing\[(4x+1)\times (4-1)=0\]
you know that 4-1 is 3 (and of course not zero.)
\[(4x+1)3=0\]
divide both sides by 3

Answer 43

i meant to type 4x-1 not 4-1

Answer 44

so
\[ (4x+1)(4x-1)=0\]
so either
4x+1=0 or 4x-1=0

WHY?
remember what I told you about a times b = zero

I am sure you can now do it!

Answer 45

\[x \pm 1/4\] ?

Answer 46

when you say x±1/4 that would mean x+1/4 or x-1/4
You menat x= ±1/4

Either way this is incorrect

Solve each one
4x+1=0 4x-1=0

Answer 47

No? 4x+1=0 and 4x-1=0, 4x=-1 and 4x=1, both sides divided by 4 are x==1/4 and x=1/4

Answer 48

I was trying to confuse you, sorry....
I do that sometimes to make sure the helper understood.
I see you did! You were correct
x= ±1/4

Answer 49

Hahaha okay... thanks!

Answer 50

so if i have the variable inside like 11(q^2-4)=0 how do i go about that?

Answer 51

you can divide both sides by 11.

Answer 52

????? can you do that out? would it just be q^2-4)=0?
sorry i keep asking for someone to do these out, i have like twenty other problems like these to do....

Answer 53

yes, it would be just q^2-4)=0
b/c


see?