algebra2sucks
What are the points of discontinuity? Are they all removable?
y = (x-3)/x^2 - 12x + 27
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myininaya
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When is the bottom 0?
myininaya
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For what values of x make the bottom 0?
algebra2sucks
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not sure?
myininaya
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So you should be able to solve an equation in this form ax^2+bx+c=0 by now.
I'm guessing this since this is a type of problem that occurs after learning that.
myininaya
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Do you know how to factor?
myininaya
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Or even use the quadratic formula to solve a quadratic equation?
myininaya
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You can even do completing the square.
algebra2sucks
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so x^2+12x+27=0?
algebra2sucks
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how do i do that?
myininaya
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Do you have -12x or +12x?
algebra2sucks
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its postive 12x
myininaya
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Hmmm...So the problem above is written/typed wrongly I assume.
Okay You should know how to factor the polynomial x^2+12x+27
myininaya
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Find two factors of c that add up to be b (you can do this when the coefficient of x^2 is 1 which in this problem it is)
myininaya
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In other words find two factors of 27 that add up to be 12.
algebra2sucks
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its like this |dw:1370027288868:dw|
myininaya
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err so it is -12x?
algebra2sucks
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noo x^2 minus 12x
myininaya
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I'm really confused what problem we are doing because twice now you have written +12x
myininaya
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You have also said twice that we have -12x
myininaya
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Anyways, we take the bottom polynomial whatever it may be and set that equal to zero and solve for x. I will let you do it since I don't know what problem we are doing anymore.
algebra2sucks
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sorry
algebra2sucks
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this is the problem were doing
myininaya
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Set bottom polynomial equal to zero.
\[x^2-12x+27=0\]
You will notice this is in the form
\[ax^2+bx+c=0\]
You have learned if you cannot factor this easily you could also use completing the square or quadratic formula.
The easy way to solve this quadratic is by factoring in my opinion.
Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.
So The factored form would be (x-m)(x-n)=0
Then you set both factors equal to 0 like so
x-m=0 or x-n=0
Solve both for x like so
x=m or x=n
algebra2sucks
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so which numbers do we use for x-m and x-n?
myininaya
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Have you found factors of 27 that have product 27 and sum -12?
myininaya
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Those are your m and n.
myininaya
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Recall this part that I said:
Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.
myininaya
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c is 27 and b is -12
myininaya
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|dw:1370027980807:dw|
myininaya
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See how I factored 27?
myininaya
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1(27)=27
3(9)=27
-1(-27)=27
-3(-9)=27
algebra2sucks
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Yes I see on my answes I have
x= -9, x =-3 is that right?
myininaya
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Do any of those factors above that multiplied to be 27 also add up to be -12?
algebra2sucks
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are these numbers removable?
algebra2sucks
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and -3?
myininaya
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(x-3)(x-9)=0
Set both equal to 0.
myininaya
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I put (x-m)(x-n)=0 where I meant (x+m)(x+n)=0.
So you would have x+m=0 or x+n=0
Which means you would have x=-m or x=-n.
But anyways
we have (x-3)(x-9)=0
And you are to set both factors equal to 0 and solve for x for both equations
myininaya
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x-3=0 or x-9=0
I leaving solving these linear equations to you.
algebra2sucks
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do we also use -1 or no?
myininaya
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Where does that come from?
myininaya
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Have you solve x-3=0 or x-9=0 yet?
algebra2sucks
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no idk how
myininaya
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x-3=0
You can't solve this for x?
You don't know how to get x by itself?
myininaya
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To undo subtraction by three you add three.
myininaya
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Try adding three to both sides and tell me what the result is.
myininaya
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x-3=0
add 3 to both sides
+3 +3
---------
what is the result?
myininaya
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x-3+3=0+3
myininaya
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what is -3+3?
what is 0+3?
algebra2sucks
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0 and 3
myininaya
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yes so
x-3+3=0+3
gives us
x+0=3
Or just x=3
myininaya
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Now you solve x-9=0
myininaya
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What are you going to add to both sides?
algebra2sucks
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9?
myininaya
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yes so x=?
algebra2sucks
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9 and 0?
algebra2sucks
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so x=9
myininaya
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yes
myininaya
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Discontinuous at x=3 or x=9
myininaya
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Now go back to the problem where you had everything factored.
myininaya
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\[\frac{x-3}{(x-3)(x-9)}\]
myininaya
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This was the problem where everything is factored, correct?
myininaya
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Does anything cancel? If so that is where you have your removable discontinuity.
myininaya
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x-3 is just a number
so is x-9
If we had 5/(5*9) you would say the 5's cancel right?
Like this 1/9
Does anything cancel?
Is there a factor on top that you see on bottom?
myininaya
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\[\frac{(x-3)}{(x-3)(x-9)}\]
If you want you can put parenthesis around the x-3 on top,
Do you see any factors that match on top and bottom?
myininaya
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This is just like matching socks.
myininaya
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I have to go to the grocery store. Good luck.
algebra2sucks
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Thanks for all ur help