Alright, this week was pretty rough for me considering i still don't have my own calcualator, but on the other hand i understood most of the stuff. One thing that was pretty tough for me were the word problems. I understood how to do most of it but when i got to the formula i didnt know what to do with it. One thing i thought was really easy was all of the different ways to find roots on anything bigger than a quadratic. The one that i will explain, that no one seems to understand is the rational root theorem.
1.) First you must find your P's and Q's.
P=All factors of the constant.
Q=All factors of the leading coefficient.
2.)Then you list all possibilities for P/Q.
3.)Next you find out which possibilites are divisible by your equation. One way to find that is to plug in your equation in your calculator, and try all possibilites. If you get zero it works. The more difficult way (which i've been doing) is to use synthetic division for all of your possibilites.
4.)Once you find a possibilite that works, you do synthetic division. The equation you get from that could be a bigger than a quadratic still. If so, you use one of your other possibilites that worked or try the same one again.
5.)Once your equation is down to a quadratic, use simple algebra, factoring, completing the square, or quadratic formula to simplify it down and get your roots.
Example:
y=x^3-4x^2+x+2
1.)P= +/-1, +/-2
Q=+/-1
Possibilites= +/-1, +/-2
2.)In this case 1 works because i used synthetic division and it worked. If you use your calculator, once you get a correct answer you then do synthetic division. Since i did already my equation is now reduced to:
x^2-3x-2
3.)Now i will use simple algebra, factoring, completing the square, or quadratic formula.
Let's see what works here.
Simple Algebra-No, because there is a middle term.
Factoring-No, because it isn't factorable.
Completing the Square-You can use it, but it would be easier to use the next form because the linear coefficient is odd, which will cause fractions.
Quadratic Formula-This is what i will use.
4.)After quadratic formula i get:
(3+(square root of 17))all over 2
(3-(square root of 17))all over 2
5.)Now I write them as points and my answers are:
((3+(square root of 17))all over 2, 0)
((3-(square root of 17))all over 2, 0)
*Sorry for all of the parenthesis, i dont know how to make square root sign, plus or minus, or exponets.*
Hope my explanation helped!!!
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I just reread my reflection and i need to have 3 answers because it is a cubic. I have no idea where i went wrong, but i'm sorry.
ReplyDeleteokay i just read amy's explanation of the rational root theorem and learned that my other point is (1,0). where that comes from is whenever i did synthetic division 1 is the number that worked so that counts as a point.
ReplyDeleteAs far as having trouble with the word problems, i think you may be getting held up on their looks. Just as a tiger looks big and bad but its really only an overgrown house cat, a word problem is just an big way to state a simple problem. Practicing word problem are a good idea not only to pass a math class but because word problems consist of real-world problem solving skills.
ReplyDeleteAny word problem has a base statement that gives you a clue to solving it.
first, read the problem two times calmly and completely. dont just get scared by their size.
then draw any picture that is described in the problem. to the side of the picture make a list of the things you know
for example you have the following word problem
"A rectangular dog pen is constructed using a barn wall as one side and 60m of fencing for the other three sides. Find the dimentions of the pen that give the greatest area "
after reading the problem you know that you need to draw the "fence line" described. therefore you need to draw a rectangle and label one long side as "barn wall"
then comes the list of things you know from the problem. dont try to guess..just take whats obviously there.
you know :
* there is 60m of fencing used on the three sides total.
* you want to find the dimentions of the fencing
therefore you need to recall the area formula
Area=Length times width or A=LxW
* you also know that because the form has two parallel sides they will be equal.
now that you have a list of things you know, put them into the picture you already have.
you need to label the two parallel sides as the same variable (X)
then since you know that there are two X's and the total fencing you have is 60m
so you can set up an equation from there!
if you have 60m and you know you need to take out 2X then your problem should read
60-2x
because this is the problem you'll use to find the last measurement of the fencing write the equation by side not labeled yet.
that concludeds setting up your problem!
its that simple.
in a word problem you dont need to see what isnt there. all you have to do is take whats given and re arrange it to make sense. everything you dont know will become clear after you organize what you do know.
now all you have to worry about is solving.
you already pulled the Area formula so you then plug in what you've written in your picture
you should have A=LxW
therefore A= X(60-2X)
now its just solving for X
first distribute the X from the outide of the parenthesis
so you should have
60X-2Xsquared
to solve from here on you can see that it’s a quadratic so you need to plug into
X= -b/2a
so you should get
X= -60/-4
which simplifies to
X= 15
this completes the hard part of the solving work.
now all that is left is plugging your X in and solving the last part of the equation
because you should have labeled your two parallel wall X's with 15
you're last equation should read
60-2(15)
then 60-30
then 30
THUS, you've solved the problem
gather your information neatly and box it off!
your final statement should read
Dimentions: 15x30