Well to start off, I think this week was pretty easy. I mean it wasn't as easy as Chapter 1, but for the most part I understood the majority of the chapter. Some things were difficult in the beginning, like memorizing the steps for the different ways to solve equations, but as I continued to practice working the same types of problems it became easier to me. I deffinitely had to study more for the Chapter 2 test yesterday than I had to for the Chapter 1 test. It was a little harder than I expected, and I had to stop and think about a few things, but I think I did alright on it.
The three things that I understood the most this week were Factoring by Grouping, Quadratic Form, and the Rational Root Theorem. They're all pretty easy because all you really have to do is know the steps for solving these types of problems.
The easiest method for me to explain is Factoring by Grouping:
1. Well first of all, the only way you can factor by grouping is if you have an even number of terms. like this problem for example because it has 4 terms>>> 8x^2 + 2x^3 + 4x + 16
2. Next, you have to make sure your terms are in order from the highest degree (exponent) to the lowest degree. so the equation would then be written as>> 2x^3 + 8x^2 + 4x +16
3. Then you have to group your terms together with parenthesis like this:
(2x^3 + 8x^2) + (4x + 16)
4. Next you look at your first set of parenthesis and ask yourself "What can I factor out?"
Since each term has at least a 2x^2 you can factor that out and you'll get>> 2x^2(x + 4)
5. Then you look at your second set of parenthesis and see what you can factor out.
Since 4 is a factor of 4 and 16, you can factor out the 4 and get>>> 4(x + 4)
6. Then when you put it all together it looks like this>> 2x^2(x + 4) + 4(x + 4)
Since you have (x + 4) twice, it cancels one of them out and you get the numbers left on the outside.>>> (2x^2 + 4) (x + 4)
7. Then you solve what's in the parenthesis by setting each one equal to zero like this:
A.) 2x^2 + 4 = 0
-4 -4
2x^2 = -4
->then divide by 2 on each side and get x^2 = -2
then square root each side and get>> x = +/- i square root of 2
B.) x + 4 = 0
-4 -4
x = -4
8. So your two answers are x = -4 and x = +/- i square root of 2.
But you have to put them in point form so your *FINAL ANSWERS* would be:
(-4,0) (i square root of 2, 0) (negative i square root of 2, 0)
**One thing I didn't quite get in Chapter 2 has to do with "Sketching Polynomial Equations." I'm not sure how to find the maximum and minimum when you have an equation with an even number of roots.
like this>> x(x-1)(x+1)(x-2)
In other words, I don't know how to find the maximum and minimum when your sketch of the polynomial is shaped like a "W"
This stumps me because when I look at the graph I don't know which max or min to find because it looks like there's two of them. By the way, I'm sorry because you probably have no idea what I'm talking about but it's really hard for me to explain it. ***But if you would like to try to help me out that would be greatly appreciated :)
Overall, I didn't have too many problems learning the new material this week and it wasn't too difficult.
Saturday, August 29, 2009
Reflection #2
this week wasnt that difficult even thought we still moved at a fast pace the material was not that difficult to understand at all. one thing i know i feel like i understand is SOLVING ANYTHING BIGGER THAN A QUADRATIC by using factoring...
1. You have to but the problem in order starting with the leading exponent to the smallest
2. then you factor out completely
3. then solve for all zero's or Xs
Ex. Find all zeros
x^3+5x^2-4x-20=0
(x^3+5x^2)-(4x+20)=0
x^2(x+5)-4(x+5)=0
(x^2-4)=0 (x+5)=0
x^2=4 x=-5
You found one of the zeros now to find the other you score root both sides...
x=+/- 2
so the answer are -2,2,-5...but then you have to put them in point form...
(-2,0)(-5,0)(2,0)...
one thing that did throw me off i think it was called the ration root theorum were you had to do long division to find all the X's im still a little confused on that...
1. You have to but the problem in order starting with the leading exponent to the smallest
2. then you factor out completely
3. then solve for all zero's or Xs
Ex. Find all zeros
x^3+5x^2-4x-20=0
(x^3+5x^2)-(4x+20)=0
x^2(x+5)-4(x+5)=0
(x^2-4)=0 (x+5)=0
x^2=4 x=-5
You found one of the zeros now to find the other you score root both sides...
x=+/- 2
so the answer are -2,2,-5...but then you have to put them in point form...
(-2,0)(-5,0)(2,0)...
one thing that did throw me off i think it was called the ration root theorum were you had to do long division to find all the X's im still a little confused on that...
Reflections #2
This week was alright, even though we still moved quick. The material wasn't that hard to understand although some parts were just a little difficult. One thing I feel like I really understand in chapter 2 is sloving anything bigger than a quadratic usuing quadratic form. To use quadratic form you have to have 3 terms only. The first term must equal the 2nd exponentx2, and the last term must be a constant. The first thing you do is make g=x^exponent/2 so that you would get g^2+g+#. The second thing is to do the quadratic formula, factor, or complete the square. The the last thing is to plug back in for g. (Whenever you do step three you are basically just plugging back into step one. g=x^2)
An example:
x^4-4x^2-12=0
1. g=x^4/2
g=x^2
g^2-4g-12
2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6
3. x^2= -2
square root of x^2=square root of -2
x=+/-i square root of 2
(i square root of 2,0)(-i square root of 2,0)
x^2=6
square root of x^2=square root of 6
x=+/-square root of 6
(square root of 6,0)(-square root of 6,0)
One thing I didn't understand that well in chapter 2 was sketching polynomial functions.
I don't get step 3 when you have to plug in on either side of your roots. And I also don't understand how to find the max and min in your calculator. And it kind of threw me off when they had and x or - in front of the problem.
An example:
x^4-4x^2-12=0
1. g=x^4/2
g=x^2
g^2-4g-12
2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6
3. x^2= -2
square root of x^2=square root of -2
x=+/-i square root of 2
(i square root of 2,0)(-i square root of 2,0)
x^2=6
square root of x^2=square root of 6
x=+/-square root of 6
(square root of 6,0)(-square root of 6,0)
One thing I didn't understand that well in chapter 2 was sketching polynomial functions.
I don't get step 3 when you have to plug in on either side of your roots. And I also don't understand how to find the max and min in your calculator. And it kind of threw me off when they had and x or - in front of the problem.
Wednesday, August 26, 2009
Help!?
Can anyone inform me on what we've been doing the past two days? I was extremely sick and I don't know what to do or how to do it. Anybody wanna help?
soo......
how do u do 9, 10, and 11 on page 89
it doesnt fit into any of the formulas that i tried
help
it doesnt fit into any of the formulas that i tried
help
help
how do u get to the thing where u plug in x's in on the calculator so that u can use the rational root theorem
Tuesday, August 25, 2009
tonights homework....HELPP
ok im not sure if im doing tonights home work right....on the first part on some of the question im coming up with zero do i use another number or just stick with that??? also when the prob. is all ready factored out but it has a negative X in front of it am i suppose to come up with only two roots or more......example #2 on 2.3... -x(x+5)(x+3) my roots are -5 and -3, what should i do about that neg. x.....on 2.4 #2 im totally lost on the prob. when it tells me to find a maxium are of the rectangle...
#2 on page 66
how do you do number two. Because when i did the second step, i came out with a zero. Is that zero a positive or a negative?
My first week(grade this,Ryan C.)
Thanks to all of those who have made my life super easy and fun my first week at riverside academy. So far things have been going great and i wish i could of had made this change much earlier. So far the school has not disappointed me yet . I also like the idea of the teacher giving us homework and giving us the option to do it to better ourselves. I find that helps motivate and push me more to actually do my homework. Anyways, this week in advance math we flew threw chapter 1 which was pretty easy. It was just alot of review things. The things we went over in chapter one was the distance formula, midpoint formula. the intersection of line, solving system of equation, slope formulas, parellel lines, complex numbers, conjugate, completing the square, and graphing parabolas. Doing the distance formula and completing the square to me was the funniest part in this chapter. They are so simple and easy to do. To solve the distance between two points just use the distance formula. It is D=the square root of(x2-x1)squared+(y2-y1)squared. Its that easy. To complete the square u must move the number to the right, divide by leading coeff, divide linear term by 2 and square it, add to both side, factor left, and then solve for x. The only thing i had a little trouble with was doin parabolas. Its not that they are hard it is just the fact that i forgot how to do the steps during the middle of the test. Surprisingly i did alright on that test. I am looking forward to week 2 when we start chapter 2.
Homework.....
Does anyone know how to work:
12) y=x^2(1-x)^2(2+x) on pg 66
AND
3) Suppose you have 102 m of fencing to make two
side-by-side rectangular enclosures, as shown.
What is the maximum area that you can enclose?
pg. 71
12) y=x^2(1-x)^2(2+x) on pg 66
AND
3) Suppose you have 102 m of fencing to make two
side-by-side rectangular enclosures, as shown.
What is the maximum area that you can enclose?
pg. 71
Monday, August 24, 2009
Help please...
Name if it is a. grouping term or b. quadratic form, then solve.
Can someone help me on #9 from homework on pg. 83:
9.) 10x^3-6x^2+5x=3
Can someone help me on #9 from homework on pg. 83:
9.) 10x^3-6x^2+5x=3
Reflection #1
I learned alot over this week but the thing i kept the most from chapter one was completing the square. It took a while for it to click in my head but i finally understood it. It turns out to be really easy too.
example-
heres the equation: 9x^2-9x+18=0
take the 9 out: x^2-x+2=0
move the 2 over: x^2-x+_____=-2
to get the next numer you divide b by 2 then square it: x^2-x+1/4=-2+1/4
simplify then factor to that vvv: x^2+x+1/4=-7/4
(x-1/2)^2=-7/4
now you just simplify to x equals till you get vvv: x-1/2= square root of -7 over 2
x= +-i square root of 7 over 2 + 1/2
example-
heres the equation: 9x^2-9x+18=0
take the 9 out: x^2-x+2=0
move the 2 over: x^2-x+_____=-2
to get the next numer you divide b by 2 then square it: x^2-x+1/4=-2+1/4
simplify then factor to that vvv: x^2+x+1/4=-7/4
(x-1/2)^2=-7/4
now you just simplify to x equals till you get vvv: x-1/2= square root of -7 over 2
x= +-i square root of 7 over 2 + 1/2
Sunday, August 23, 2009
Reflection #1
Im having trouble with number 5 on the chapter 1 test. Now i think i know what im doing wrong but i was wondering if someone could tell me if im right.
The problem says to find the y intercept of: y=x^2+6x+4
I start off by using the quadratic formula and come to: negative six plus or minus squareroot of 36 minus 16 all over 2.
Which simplifies to: negative 3 plus/minus squareroot of 5
Now this is where i believe i messed up
Was i supposed to factor that back into the problem because i think i found the x-intercepts?
The problem says to find the y intercept of: y=x^2+6x+4
I start off by using the quadratic formula and come to: negative six plus or minus squareroot of 36 minus 16 all over 2.
Which simplifies to: negative 3 plus/minus squareroot of 5
Now this is where i believe i messed up
Was i supposed to factor that back into the problem because i think i found the x-intercepts?
Reflection 1
This week we learned about the midpoint formula, Distance Formula, Solving a System of Eqns, Imaginary Numbers chart, quadratics (Completing squares, Factoring, Quadratic Formula), Graphing Parabolas, and Polynomials. The easiest that came to me was synthetic division. Back in Alg. II, i thought synthetic division was stupid and pointless and hard. But this week i actually understood how to do it and realized it's simpler than i thought. What i dont understand is graphing parabolas. I dont understand how to find the x and y intercepts or the axis of symmetry. Somebody needs to help me out with that...lawl.
Answer to #19 on Chapter 1 Test
Hey everybody,
I noticed a lot of you are having trouble this #19 also. Here's the problem.
Find the equation of the perpendicular bisector of the line 4x - 2y = 4 through the point (1, 4).
Here's how it works. There is a specific formula used for this kind of problem. It is called point-slope form. This is the formula: y - y1 = m (x -x1)
So, what you do is solve for y.
4x - 2y = 4
-2y = -4x +4
y = 2x -2
Now you have to find the parallel slope of this line.
The slope is 2. To find the parallel slope, you take the negative reciprocal of the slope.
So, 2 would become -1/2.
So, now you just plug in the point into the formula along with the parallel slope, (m = parallel slope in this particular problem).
y - y1 = m (x - x1)
(1, 4)............y1 = 4..........x1 = 1
y - 4 = -1/2 (x - 1)
That's your answer.
Reflection #1
So yeah, although it was mainly review, left me with the "deer in headlights" look and my mouth wide open. On the other hand, surprisingly there were many concepts that i am completely comfortable with now. Factoring and quadratic formula feel like second-nature to me now:
the easiest equations to factor are the equations where the leading coefficient is 1, i.e:
x^2+6x-7=0
Obviously, from this step in order to factor this equation; the factors of the last number must add or subtract to equal the linear term.
(x+7) (x-1)=0
By setting the two equations equal to 0, you can find your two answers.
x=-7,1
to finish the equation put it in point form:
(-7,0) (1,0)
For quadratic formula,
i find it easiest to remember the formula if you say it a few times fast, kind of like a tongue twister! (i know i'm weird, but i haven't forgotten it since i first learned it.)
opposite of b, plus or minus the square root of b minus a times c all over 2 times a.
first, define a,b,c.
second, plug them into your formula.
third, this equation should give you two answers.
(if you don't have two, you did something wrong)
fourth put your two answers in a+bi form, and also in point form.
---------------------------------------------------------------------------
hm, to say the least, i had no earthly idea how to work number 19 on chapter 1 test.
Did anyone get this problem!?
Find the equation of the equation of the perpendicular bisector of the line 4x-2y=4 through the point (1,4).
i did simple algebra and came out completely wrong, i plugged in the y for y and x for x.
maybe i should have solved for y and plugged it in and solved for x?
4x-2y=4
-2y=4-4x
y=-2+2x
4x-2(-2+2x)=4
4x+4-4x=4
4=4?
is that possible?!
or maybe i should have plugged in the given y coordinate for y and solved then plugged back in for x?!
4x-2(4)=4
x=3
ah, so many ways to mess up. lol
if anyone has a clue, i'd greatly appreciate some help!
please and thanks! :)
the easiest equations to factor are the equations where the leading coefficient is 1, i.e:
x^2+6x-7=0
Obviously, from this step in order to factor this equation; the factors of the last number must add or subtract to equal the linear term.
(x+7) (x-1)=0
By setting the two equations equal to 0, you can find your two answers.
x=-7,1
to finish the equation put it in point form:
(-7,0) (1,0)
For quadratic formula,
i find it easiest to remember the formula if you say it a few times fast, kind of like a tongue twister! (i know i'm weird, but i haven't forgotten it since i first learned it.)
opposite of b, plus or minus the square root of b minus a times c all over 2 times a.
first, define a,b,c.
second, plug them into your formula.
third, this equation should give you two answers.
(if you don't have two, you did something wrong)
fourth put your two answers in a+bi form, and also in point form.
---------------------------------------------------------------------------
hm, to say the least, i had no earthly idea how to work number 19 on chapter 1 test.
Did anyone get this problem!?
Find the equation of the equation of the perpendicular bisector of the line 4x-2y=4 through the point (1,4).
i did simple algebra and came out completely wrong, i plugged in the y for y and x for x.
maybe i should have solved for y and plugged it in and solved for x?
4x-2y=4
-2y=4-4x
y=-2+2x
4x-2(-2+2x)=4
4x+4-4x=4
4=4?
is that possible?!
or maybe i should have plugged in the given y coordinate for y and solved then plugged back in for x?!
4x-2(4)=4
x=3
ah, so many ways to mess up. lol
if anyone has a clue, i'd greatly appreciate some help!
please and thanks! :)
Reflection #1
Well yeah, its chad, and i guess im coming say what i learned and stuff. I learned alot over this week but the thing i kept the most from chapter one was completing the square. It took a while for it to click in my head but i finally understood it. It turns out to be really easy too.
example-
heres the equation: 9x^2-9x+18=0
take the 9 out: x^2-x+2=0
move the 2 over: x^2-x+_____=-2
to get the next numer you divide b by 2 then square it: x^2-x+1/4=-2+1/4
simplify then factor to that vvv: x^2+x+1/4=-7/4
(x-1/2)^2=-7/4
now you just simplify to x equals till you get vvv: x-1/2= square root of -7 over 2
x= +-i square root of 7 over 2 + 1/2
i dont know if anyone else got a little confused on this but i did no doubt.
Reflection # 1
Going into this week of advance math I was quite nervous, because you had told us that we were having a chapter 1 test that up coming Friday. I got a little confused with all the different formulas brob threw at us. On the test I wasn't sure when to apply certain formulas, especially with graphing parabolas. I probally should have took more time to look over all the steps the night before the test, but one thing i did understand is how to use and follow was the i chart. This is where you take the exponents of i and divide the number by 4 then all you have to do is know the chart. Number 16 on the test is where you use this concept.
This is the i chart: .25 = i
.5 = -1
.75 = -i
1or any whole number = 1
Example: i^498+i^185+i^2003+i^16+i^89=
So first step you divide each exponent by 4: i^498=124.5, i^185=46.25, i^2003=500.75, i^16=4, i^89= 22.25
Next step look at the decimals or whole numbers you got and plug them into the chart and next into the equation : (-1) + i + (-i) + 1 + i =
Solve: The two one's cancel out and two of the i's also cancel out so you left with i which is your answer.
_______________________________________________________________________________
On the other hand I do not have a clue how to approach number 19 on the chapter 1 test:
Find the equation of the equation of the perpendicular bisector of the line 4x-2y=4 through the point (1,4).
Im guessing that you have to put it in point slope form or something like that.
This is the i chart: .25 = i
.5 = -1
.75 = -i
1or any whole number = 1
Example: i^498+i^185+i^2003+i^16+i^89=
So first step you divide each exponent by 4: i^498=124.5, i^185=46.25, i^2003=500.75, i^16=4, i^89= 22.25
Next step look at the decimals or whole numbers you got and plug them into the chart and next into the equation : (-1) + i + (-i) + 1 + i =
Solve: The two one's cancel out and two of the i's also cancel out so you left with i which is your answer.
_______________________________________________________________________________
On the other hand I do not have a clue how to approach number 19 on the chapter 1 test:
Find the equation of the equation of the perpendicular bisector of the line 4x-2y=4 through the point (1,4).
Im guessing that you have to put it in point slope form or something like that.
Reflection # 1
First, this week was a lot rougher that what I was expecting. We flew through Chapter 1 faster than I could blink my eyes. I was not expecting this week to go this fast but if I liked it or not it did. With all the formulas and homework it was definitly a wake up call to stop thinking about summer and start on school. The things that I found the easiest are quadratic formula, and completing the square. Here’s an example of quadratic formula to show how it works:
3x^2-4x+1=0
-b+-square root b^2-4ac/2a
--4+-square root (-4) ^2-4(3) (1)/2a
4+- square root 4/6
4+-2/6
6/6=1 (1,0)
2/6=1/3 (1/3,0)
Step 1: take the equation and plug it into the formula.
Step 2: plug everything under the square root into your calculator.
Step 3: simplify the square root then simplify the fraction and plug in as a point.
The thing I had the most trouble with is Graphing Parabolas. I understand everything till you have to solve for the vertex. In the example solving for the vertex, the original problem is y=2x^2-8x+5. The example it has (2,-3), 2(2)^2-8(2)+5=-3 so where does the 2 come from? Does it matter what number you plug in?
As the school year goes on and we learn more it will get tougher but we will all get through it.
3x^2-4x+1=0
-b+-square root b^2-4ac/2a
--4+-square root (-4) ^2-4(3) (1)/2a
4+- square root 4/6
4+-2/6
6/6=1 (1,0)
2/6=1/3 (1/3,0)
Step 1: take the equation and plug it into the formula.
Step 2: plug everything under the square root into your calculator.
Step 3: simplify the square root then simplify the fraction and plug in as a point.
The thing I had the most trouble with is Graphing Parabolas. I understand everything till you have to solve for the vertex. In the example solving for the vertex, the original problem is y=2x^2-8x+5. The example it has (2,-3), 2(2)^2-8(2)+5=-3 so where does the 2 come from? Does it matter what number you plug in?
As the school year goes on and we learn more it will get tougher but we will all get through it.
Reflections 1, reattempt
So, to start off, this week wasnt as hard as i thought it would be. I mean, the test was disliked, but meh, watcha gonna do. It went by really fast, but at least its over. Except for this blog, the week was a breeze. See, if my a.d.d. didnt kick in, then youd be seein a big blog, but no, it decides to kick in right when im doin somethin that i need for school. And its hard to remember stuff that happened last week right off the top of my head, but i do remember that it wasnt really hard, just remembering how to find certain points and equations got to me last week. There were many problems that messed me up on the homework, but that was because i misunderstood the formulas and such, so it was my fault. But now, i understand them, so no worries on that part.
So now, i have no idea what to put for the next 100 or so words in this blog, except what i do and dont understand. But meh, gotta fill in the words, so the rest is gonna be stuff about problems and ranting if theres not enough words.
So one thing i do understand is the distance formula. I know its dumb to state that on here, but i felt like posting somethin easy that i understood on here.
Example: find the distance between point (4,3) and point (2,6)
first u gotta plug the points into the formula D=√((x₂-x₁)^2+(y₂-y₁)^2)
then u use basic math to find the outcome
D=√((2-4)^2-(6-3)^2)
D=√((-2)^2-(3)^2)
D=√(4-9)
D=√-5
then u have to use i infront of the √ sign because u cant have the square root of a negative number
D=±i√5 is ur final answer
And something that i dont understand, and need help with, is #19 on the chapter 1 test. I dont understand how you find the equation of a bisector of the line, it confuses me with points and how it has to be through a certain point. I have no idea how to get a line to go through a certain point when im only given certain numbers, its not enough.
So now, i have no idea what to put for the next 100 or so words in this blog, except what i do and dont understand. But meh, gotta fill in the words, so the rest is gonna be stuff about problems and ranting if theres not enough words.
So one thing i do understand is the distance formula. I know its dumb to state that on here, but i felt like posting somethin easy that i understood on here.
Example: find the distance between point (4,3) and point (2,6)
first u gotta plug the points into the formula D=√((x₂-x₁)^2+(y₂-y₁)^2)
then u use basic math to find the outcome
D=√((2-4)^2-(6-3)^2)
D=√((-2)^2-(3)^2)
D=√(4-9)
D=√-5
then u have to use i infront of the √ sign because u cant have the square root of a negative number
D=±i√5 is ur final answer
And something that i dont understand, and need help with, is #19 on the chapter 1 test. I dont understand how you find the equation of a bisector of the line, it confuses me with points and how it has to be through a certain point. I have no idea how to get a line to go through a certain point when im only given certain numbers, its not enough.
reflection #1
this week went by pretty fast, but it wasnt too bad for the first week of class. one of the things i retained from chapter one (after looking back at it) was finding an intersection. this is how you do it...
eg: y=2x2-8x+5
step 1. does it open up or down?
if its positive it opens up, and if its negative it opens down
step 2. how many x-intercepts does it have?
b2-4ac discriminate: positive-2x intercepts negative-no x intercept 0-1 x intercept
(-8)2 -4(2)(5)=24 so there are 2 x intercepts
step 3. find the intercepts
complete the square to find
2x2-8x+5
2x2-8x =-5
x2-4x+4=-5/2+4
(x-2)2=3/2 [shortcut]
x-2= [square root of] 3/2
x-2= [square root] 3/ [square root] 2 X [square root] 2/ [square root]2
x=2+/- [square root] 6/2
put in point form
(2+ [square root]6/2, 0) (2- [square root] 6/2, 0)
step 4. find the y-intercept
plug in y
2(0)2-8(0)+5
=5
(0,5)
step 5. find the axis of symmetry
x=-b/2a
=-(-8)/2(2)
=2
step 6. find the vertex
x=2
2(2)2-8(2)+5
=-3
vertex= (2, -3)
reflections 1
This week wasn't too bad. I probably grasped the concept of factoring better than anything else tho, so i'll explain that.
Ok so to start you have to start you look at a quadratic equation, and it needs to be set equal to 0. There should be an x^2 term, an x term, and a linear term, in that order. To factor, multiply the x^2 coefficient by the linear term:
2x^2+7x+5
2 x 5=10
Now you find factors of 10 that add to give you 7.
5 and 2 add to give you 7 so now put it in this form: (2x^2+5x)+(2x+5)
-notice where the 2 and the 5 plug in-----------------------^----^----------
-now we factor this out: remove an x from the 1st group= x(2x+5)
-now we have: x(2x+5)+(2x+5)=0
-x=0------------^-----------------^----------
-now we simplify 2x+5=0, which is x= -5/2.
So the answer to the factor problem is (0,0) and (-5/2,0) because it needs to be in point form.
If the equation has no x^2 coeffictient:
-Take factors of the linear term that add to give you the x coefficient:
x^2+3x+2
2 x 1=3
So now (x+2)+(x+1)
Answer is (-2,0) (-1,0)
_____________________________________________________________________
On the other side of the rainbow, I didn't quite grasp the concept of finding the equation of a perpendicular bisector of a line. Like in a case of the test where you have to find the perpendicular bisector of the line 4x-2y=4
-I know that first you have to solve the equation for y.
-Then I know that perpendicular slope is -1/2.
B-Rob put that the answer was y-4=-1/2(x-1)
-I understand that that is in point slope form, but I don't know where the y-4 or the (x+1) came from.
Ok so to start you have to start you look at a quadratic equation, and it needs to be set equal to 0. There should be an x^2 term, an x term, and a linear term, in that order. To factor, multiply the x^2 coefficient by the linear term:
2x^2+7x+5
2 x 5=10
Now you find factors of 10 that add to give you 7.
5 and 2 add to give you 7 so now put it in this form: (2x^2+5x)+(2x+5)
-notice where the 2 and the 5 plug in-----------------------^----^----------
-now we factor this out: remove an x from the 1st group= x(2x+5)
-now we have: x(2x+5)+(2x+5)=0
-x=0------------^-----------------^----------
-now we simplify 2x+5=0, which is x= -5/2.
So the answer to the factor problem is (0,0) and (-5/2,0) because it needs to be in point form.
If the equation has no x^2 coeffictient:
-Take factors of the linear term that add to give you the x coefficient:
x^2+3x+2
2 x 1=3
So now (x+2)+(x+1)
Answer is (-2,0) (-1,0)
_____________________________________________________________________
On the other side of the rainbow, I didn't quite grasp the concept of finding the equation of a perpendicular bisector of a line. Like in a case of the test where you have to find the perpendicular bisector of the line 4x-2y=4
-I know that first you have to solve the equation for y.
-Then I know that perpendicular slope is -1/2.
B-Rob put that the answer was y-4=-1/2(x-1)
-I understand that that is in point slope form, but I don't know where the y-4 or the (x+1) came from.
Reflection #1
In chapter one I understood completing the sqare..
First, you must make sure that your xsquared term has no number in front of it. You must divide by the number in fron of it if there is one. Then take your "c" term and make sure its on the right side of the =. Then divide your X term by 2 and square it and add it to both sides. Then you would take your number that you used to square it and add it to the problem and solve the right side of the problem.
EXAMPLE:
x2-4x=9
x-4x =9
(-4/2)=(-2)
(-2)2 = 4
x-4x+4=9+4
(x-2)2=13
then you would take the square root from both sides
x-2=(square root)13
13 has no perfect square root..
then you would add 2 to both sides and end up with
x= 2+or-(the square root of)13
and put it in point form for b.rob :)
on the other hand I completely do not understand how to solve number 16 on free response on the test:
16.) Simplify i2453 + i1998 + i2006
reflections 1
i really hope i'm posting this in the right place. i just finished correcting my test and i don't understand how i missed all of those questions, but there is one that i'm still have trouble with number 19. Find the equation of the perpendicular bisector of the line 4x-2y=4 everytime i read that question i just draw a blank. so if anyone know's what my problem is feel free to help.
this week went by super fast and surprisingly i felt we all kept up pretty well, i also feel that calculators are extremely expensive! ;) okay anyways the one thing i understood this week was axis of symmetry
example number 6 on test y=2x^2-4x+3
all you do is plug it into your function x=-b/2a
this is your abc of the equation above.
a- 2
b- (-4)
c- 3
so plug it in:
x= -(-4)/2(2)
which then gives you
x= 4/2(2)
solve the denom.
x=4/4
and since 4/4 equals 1
your answer is
x= 1
its really simple you just need to remember your function for that problem.
this week went by super fast and surprisingly i felt we all kept up pretty well, i also feel that calculators are extremely expensive! ;) okay anyways the one thing i understood this week was axis of symmetry
example number 6 on test y=2x^2-4x+3
all you do is plug it into your function x=-b/2a
this is your abc of the equation above.
a- 2
b- (-4)
c- 3
so plug it in:
x= -(-4)/2(2)
which then gives you
x= 4/2(2)
solve the denom.
x=4/4
and since 4/4 equals 1
your answer is
x= 1
its really simple you just need to remember your function for that problem.
Answer to #18 on Chapter 1 Test
Hey everybody,
I know a lot of you are having trouble figuring out this problem on the test, so I have the solution for you guys.
So it says to find the reciprocal of i and simplify.
The reciprocal of i is 1 over i, or 1/i. Since you can't have i in the denominator, you must multiply both the numerator and denominator of the fraction by i, or -1.
So 1/i · i/i = i/i².
So i/i² simplifies to i/-1, because i² = -1.
Now you must multiply i/-1 by -1/-1, which gives you your solution:
-i/1, which translates simply to....
-i
Reflection # 1
This week along with Chapter 1 was very easy. I understood most of the content including the distance and midpoint formulas, the systems of equations, solving for x-intercepts, and graphing parabolas (which was probably the most fun learning, applying, and understanding).
So one thing I completely understood from this week was solving for the x-intercepts, which are also called roots, or zeroes, in an equation. But I will specifically describe the method of solving for x-intercepts in the way of the quadratic formula.
Example problem:
2x² - x - 6 = 0
The formula for solving for x is (-b ± √b² - 4ac) / (2a)
(So this means take the negative of the coefficient of x, plus or minus, the square root of b² minus 4 times the coefficient of x² times the constant. Then divide that by 2 times a.
So for this problem we have:
2x² - x - 6 = 0
(-(-1) ± √(-1)² - 4 (2) (-6)) / (2(2))
(1 ± √1 - (-48)) / 4
1 ± √49 / 4
1 ± 7 / 4
which means.....
x = 1 ± 7 / 4
x = 8 / 4 and x = -6 / 4
x = 2 and x = -3 / 2
Now put the answers in point form.
This is your answer.
(2, 0) and (-3/2, 0)
Okay, now one thing I didn't understand was #24 in Friday's homework on page 57.
If 2i is a zero of f(x) = x⁴ + x² + a, find the value of a.
Would this deal with synthetic division? Would you divide this by 2i? Can somebody please explain? Thanks.
Reflection #1
this whole week has been kind of crazy. between the rapid reviewing and trying to grasp a hold of the formulas, i started to get a little confused at times because i started to forget whcih formulas went with thre right problem. there was somithing ireally did not understand. how to finda an equation of a perpendicular bisector and how to find a parallel slope of a line.
Examples are number 2 and 19 on the chapter 1 test. but other than that i feel confident aout everything esle.
Completing the Square
when completng the square the first thing you have to di is put all X's to one side of the equation and your whole numbers to the other side of the equation. then you make your quadratic term equal to 1(it should be x^2). then you take your binomial co eff. and pit it in this formula (-b/2(a)^2. you multiply your quadratic co eff. by two and then you divide. after you come up with the first answer you square it. then you plug the answer into the problem on both sides of the equation x^2+x+1=3+1. then you simplify. ageter that you take the answer you had before and you square it and put it in another equation and it would be (x+1)^2=4. the you square root bothe sides and solve for x and your final answer would be x=3.
Examples are number 2 and 19 on the chapter 1 test. but other than that i feel confident aout everything esle.
Completing the Square
when completng the square the first thing you have to di is put all X's to one side of the equation and your whole numbers to the other side of the equation. then you make your quadratic term equal to 1(it should be x^2). then you take your binomial co eff. and pit it in this formula (-b/2(a)^2. you multiply your quadratic co eff. by two and then you divide. after you come up with the first answer you square it. then you plug the answer into the problem on both sides of the equation x^2+x+1=3+1. then you simplify. ageter that you take the answer you had before and you square it and put it in another equation and it would be (x+1)^2=4. the you square root bothe sides and solve for x and your final answer would be x=3.
Reflection #1
Okay, chapter 1 was super simple. One thing I completely understand and can explain is completing the square: (To make it easier, I put it into steps.)
Ex. x^2+x-2=0
Step 1: Move c over x^2+x =2
Step 2: Divide by quadratic coefficient x^2+x =2
Step 3: Divide b by 2, the square it and add to both sides (1/2)^2=1/4 x^2+x+1/4=2+1/4
Step 4: Factor (x+1/2)^2=9/4
Step 5: Square root both sides x+1/2=+/-3/2
Step 6: Solve for x x=-1/2+/-3/2 x=1 x=-2
Step 7: put in point form (#,#) (1,0) (-2,0)
Now, I don't understand #18 on the Ch. 1 test. It says "Find the reciprocal of i and simplify". I know what the reciprocal of i is, but I'm not sure how the simplify it. I think you rationalize it, but... I don't know, but could someone explain it to me? Please and thank you!
Ex. x^2+x-2=0
Step 1: Move c over x^2+x =2
Step 2: Divide by quadratic coefficient x^2+x =2
Step 3: Divide b by 2, the square it and add to both sides (1/2)^2=1/4 x^2+x+1/4=2+1/4
Step 4: Factor (x+1/2)^2=9/4
Step 5: Square root both sides x+1/2=+/-3/2
Step 6: Solve for x x=-1/2+/-3/2 x=1 x=-2
Step 7: put in point form (#,#) (1,0) (-2,0)
Now, I don't understand #18 on the Ch. 1 test. It says "Find the reciprocal of i and simplify". I know what the reciprocal of i is, but I'm not sure how the simplify it. I think you rationalize it, but... I don't know, but could someone explain it to me? Please and thank you!
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