Math is my worst subject; thus, I've been strugging with these questions for the past few hours and hae asked various people but most haven't done this in years so I have ventured to the Yoni forum to ask. :P I'm doing homework (study guide) and studying for my Algebra II test tommorow. I have some questions (even though it says it's not for homework help.)
1) I've noticed that in a few example problems my teacher did while getting factors from zeros of a paraballa's(sp?) curve that they have been the exact opposite. (example: Zeros of the function are (-1,0) & (5,0) while the factors are (X-5) & (X+1) is this always the case?
2)What is the Square root method and what's the format for it? Example: (the question from the homework)
Quotey=-x^2+9
Find the roots (aka zeros/x-intercept) using the square root method
3)Find the roots (aka zeros/x-intercept) by factoring completely.
Sample problem:
y=x^2-10x+24
4)Directions: Use the quadratic formula to find the roots. Leave answers in simplified radical form.
Sample Problem:
y=x^2+4x+12
5) Looking back at my first question, how can you use factors to write the quadratic equation in standard form? (example: Factors are (x-2) and (x+4) )
These questions below come directly from the optional study guide - I don't get graded or checked on this but the questions above were about the ones I actually do get graded on. I've done all but the problems that have that info on it.
QuoteUse the following equations to answer questions 13-16.
a) 2x^2 - 4x + 16 b) 4x^2 - 4x + 1
c) 4x^2 - 81 d) 4x^2 + 81
Questions:
13) Which equation is a perfect square trinomial? (I'm guessing here but...B?)
14) Factor the perfect square trinomial.
15) Which equation is the difference of squares? ...what?
16) Factor the difference of squares? uh..so what's difference of squares mean?
Thanks for any help
1. yes.
2. possibly y = 0 = -x^2+9 => x = +/- sqrt(9) = +/- 3
3. x^2-10x+24 = (x+a)^2 + b = x^2 + 2ax + a^2 + b
=> 2a = -10 => a = -5
a^2 + b = 24 = (-5)^2 + b => b = -1
so, x^2 -10x + 24 = (x-5)^2 -1
Now find the zeros, set (x-5)^2 - 1 = 0 and use square root to find that (x-5) = +/-1, x = 5 +/-1, x1 = 4, x2 = 6
This means that the factored form is (x-4)(x-6), also verify that (x-4)(x-6) = x^2 - 4x - 6x + 24 = y
4. I'm not at all sure what quadratic formula is or simple radical form, but I'd guess that they refer to the generic formula found by
x^2 + px + q = (x + a)^2 + b
again identifying
2a = p, b = q - a^2, a = p/2, b = q - p^2/4
then set to zero and solve using square root:
x^2 + px + q = 0 = (x + a) ^ 2 + b =>
(x+a)^2 = -b => x = +/- sqrt(-b) - a = +/-sqrt(q - p^2/4) - p/2
That formula should be the quadratic formula which you might memorize.
For y=x^2+4x+12, p = 4 and q = 12, so
x = -2 +/- sqrt(12 - 4^2/4) = -2 +/- sqrt(8) = -2 +/- 2 sqrt(2)
I'll assume that radical form includes sqrt's.
5. I don't understand
13. I think a perfect square trinomial has no b term. For b to be zero, q has to be p^2/4.
Trying that on b) after dividing by 4 to leave x^2 without a factor in front of it gives p = -1, q = 1/4 = (-1)^2/4, so b is indeed zero for that.
14. This means that the polynom has two equal roots, x = - p/2 = -1/2, and so 4x^2 - 4x + 1 = 4(x - 1/2)^2 = (2x - 1)^2
15. Difference of squares is probably for the special case (a+b)(a-b) = a^2 - b^2. Just look for two squares subtracted from each other - has to be c).
16. Factoring a difference of squares is easy, use the well known formula to immediately get 4x^2 - 81 = (2x + 9) (2x - 9)
Quadratic Formula Is....
Quote
-b +/- sqrt(b^2-4ac)
x= ----------------------------
2a
A, B, C is in reference to..
Quote
ax^2 + bx + c = 0
Thanks for your help Adron, I haven't checked 2-4 yet however, thanks for clearing up #1.
Edit: added quote tags
For more detail on #1, the idea is that if you have two numbers multiplied by each other, the result will only be zero if at least one of the numbers is zero.
Now, if you multiply (x-a) and (x-b) and get a zero result (the function is crossing the x axis), you know that either x-a is zero or x-b is zero (or both).
For x-a to be zero, x has to be equal to a, and for x-b to be zero, x has to be equal to b. Thus if you know the zeroes, you know how to write the factored function, and reverse.
You use a more complex quadratic formula than what I use. My first step is always to divide by any constant in front of x^2, so I only have to worry about the two variables p and q when factoring. The constant can always be added back later.
Alrighty.
Quote5) I don't understand
This is the problem as I have it...
QuoteA quadratic equation has factors (x-2) and (x+4).
Use the factors to write the quadratic equation in standard form.
As far as I got was that the zeros were (2,0) and (-4,0)
QuoteA quadratic equation has factors (x-2) and (x+4).
Use the factors to write the quadratic equation in standard form.
Here you use the FOIL method.
First
Outside
Inside
Last
(x - 2) * (x + 4)
Multiply the first term inside each parenthesis by eachother:
x * x = x ^ 2:
The outside term * the outside term:
x * 4 = 4x;
Inside term * Inside term:
-2 * x = -2x
Last term * Last term:
-2 * 4 = -8
Add them all together:
x^2 + 4x - 2x - 8 =
x^2 + 2x - 8
There's your answer in standard form.
I was taught the 'rainbow' method as opposed to FOIL. I hate teachers.
Alright, thanks all. I think I passed the test, I guess I'll find out when I come back from break. I came home sick so I just stayed long enough to take it. Thanks for the help I would of gotten a few questions wrong had I not asked.
Hi, sorry for missing this thread.
Quote from: hismajesty on December 17, 2003, 06:47 PM
I have some questions (even though it says it's not for homework help.)
I didn't say it's not for homework help, just that I won't do your homework. :)
I can help you with it, but I won't do it for you without you learning anything.
Ah, alright thanks.
Quote from: Orillion on December 17, 2003, 11:17 PM
I was taught the 'rainbow' method as opposed to FOIL. I hate teachers.
What is this rainbow method?
(http://www.templatecentral.net/rainbow.JPG)
?
FOIL Transitivity. Same thing.
I never understood the American FOIL and similar tricks. We have no such thing. It couldn't be much simpler, just multiply everything by everything else and add all the pairs.
(a+b)(c+d) = ac + ad + bc + bd
What's the big deal?
Quote from: Yoni on December 18, 2003, 04:36 PM
I never understood the American FOIL and similar tricks. We have no such thing. It couldn't be much simpler, just multiply everything by everything else and add all the pairs.
(a+b)(c+d) = ac + ad + bc + bd
What's the big deal?
Little tricks are less confusing for us 'stupid' Americaniums. :P
So how do americans multiply
[a b c] [p]
[x y z] * [d e f] * [q]
[g h i] [r]
? What's your trick?
Quote from: Adron on December 18, 2003, 05:41 PM
So how do americans multiply
[a b c] [p]
[x y z] * [d e f] * [q]
[g h i] [r]
? What's your trick?
TI-83+
And what does that say about the problem?
Quote from: Adron on December 18, 2003, 06:01 PM
And what does that say about the problem?
That it's a matrix multiplication problem?
I thought you'd be able to figure that out without asking your TI-83+?
:P
Quote from: Adron on December 18, 2003, 06:56 PM
I thought you'd be able to figure that out without asking your TI-83+?
Americans must have super calculators now days which requires students to do no thinking.
Quote from: Spht on December 18, 2003, 09:36 PM
Quote from: Adron on December 18, 2003, 06:56 PM
I thought you'd be able to figure that out without asking your TI-83+?
Americans must have super calculators now days which requires students to do no thinking.
We have to learn how to turn them on.
Ugh you TI-83+ dummies. Adron showed a first-degree two-vector polynomial in matrix form. You should recognize that for what it is:
axp + bxq + cxr + dyp + eyq + fyr + gzp + hzq + izr
Edit: "Two-vector" instead of "two-variable", there are 6 variables.
Hehe, that was exactly what I wanted him to show :)
Complain to my teacher, she said you'll never have to deal with that without using the calculator. :-[
Quote from: hismajesty on December 19, 2003, 08:14 AM
Complain to my teacher, she said you'll never have to deal with that without using the calculator. :-[
She must not think very highly of her students... :(
Ah but what gets me is that we can't use calculators on the test. So instead of teaching us the way to do it without one she didn't include it on the test earlier this year. :-\
Quote from: hismajesty on December 17, 2003, 06:47 PM
1) I've noticed that in a few example problems my teacher did while getting factors from zeros of a paraballa's(sp?) curve that they have been the exact opposite. (example: Zeros of the function are (-1,0) & (5,0) while the factors are (X-5) & (X+1) is this always the case?
Yeah, I'm a little late, but anyway, the reason it's like this is the zeroes are what you have to add to the factor to get 0.
So for the two examples, you would do this:
X - 5 = 0 --> x = 5
X + 1 = 0--> x = -1
Make sense?
Quote from: iago on December 19, 2003, 10:18 PM
Quote from: hismajesty on December 17, 2003, 06:47 PM
1) I've noticed that in a few example problems my teacher did while getting factors from zeros of a paraballa's(sp?) curve that they have been the exact opposite. (example: Zeros of the function are (-1,0) & (5,0) while the factors are (X-5) & (X+1) is this always the case?
Yeah, I'm a little late, but anyway, the reason it's like this is the zeroes are what you have to add to the factor to get 0.
So for the two examples, you would do this:
X - 5 = 0 --> x = 5
X + 1 = 0--> x = -1
Make sense?
That works but actually it's because in the general form of the equation, the sign is negative, so all you have to do is reverse the sign. So (x-h) and (y-k) are the general form, and if you have (x-5) and (y+1) the 5 is positive, and the 1 is negative because a negative and a negative make a positive.
But it's like that because you're trying to make them equal to zero.
Quote from: Yoni on December 18, 2003, 04:36 PM
I never understood the American FOIL and similar tricks. We have no such thing. It couldn't be much simpler, just multiply everything by everything else and add all the pairs.
(a+b)(c+d) = ac + ad + bc + bd
What's the big deal?
Hehe. When I expand I usually do the following:
(a+b)(c+d) = ac + bc + ad + bd
FIOL!
Edit: I think the reason we were taught this method is that it allows us to give the method a name when explaining. The same can be said for the rainbow method. "Rainbow it out" or "Foil it out".
The American educational system, nevermind simply refering to mathmatics in general, is totally fucked up and it would seem that the only good teachers are those in colleges (and I've met a few that are pretty dumb though) and unless you have tons of money, you really won't get much out of public education opposed to those obviously in other countries unless your willing to learn on your own.
The math teachers I had in HS were fucking idiots and taught us stuff, but didn't do it efficiently because most kids were just as dumb if not dumber when they left the class. When it came to using calculators, they were depended upon in classes and I remember some classes where we spent the entire time learning how to use a TI-83 and how to manipulate the graphing functions. Then most of the time as stated before, we weren't allowed to use the calculators but had to rely on what we had supposively been taught.
I remember taking my "Intermediate Algebra" class last spring with one of my favorite math teachers and I'll never forget losing my patients with him when he was teaching us logorithms and natural logs. I had asked him HOW logs worked because of course like most programmers, I want to know how people did logs BEFORE calculators and other bits of technology were around. Of course he didn't understand my question I don't think because I believe he thought I wanted a programatic version of the algorithm instead of explaining it in class which was false. I just think he thought the students wouldn't really care about the actual method.
As a last note about using calculators in math class, we used to play nes games on ours and I loved playing mario on mine :)
I have mario on mine but hardly play it, I usually play pacman. Infact the TI-83 is what got me in to programming way back in 7th grade.
Quote from: DarkVirus on December 21, 2003, 10:13 PM
The American educational system, nevermind simply refering to mathmatics in general, is totally fucked up and it would seem that the only good teachers are those in colleges (and I've met a few that are pretty dumb though) and unless you have tons of money, you really won't get much out of public education opposed to those obviously in other countries unless your willing to learn on your own.
The math teachers I had in HS were fucking idiots and taught us stuff, but didn't do it efficiently because most kids were just as dumb if not dumber when they left the class. When it came to using calculators, they were depended upon in classes and I remember some classes where we spent the entire time learning how to use a TI-83 and how to manipulate the graphing functions. Then most of the time as stated before, we weren't allowed to use the calculators but had to rely on what we had supposively been taught.
I remember taking my "Intermediate Algebra" class last spring with one of my favorite math teachers and I'll never forget losing my patients with him when he was teaching us logorithms and natural logs. I had asked him HOW logs worked because of course like most programmers, I want to know how people did logs BEFORE calculators and other bits of technology were around. Of course he didn't understand my question I don't think because I believe he thought I wanted a programatic version of the algorithm instead of explaining it in class which was false. I just think he thought the students wouldn't really care about the actual method.
As a last note about using calculators in math class, we used to play nes games on ours and I loved playing mario on mine :)
None of my math classes save pre-calculus (Which was a complete waste of time) were dependant upon calculators...and this was a public high school. If anything, the whole math department shunned the use of calculators.
throughout highschool, I did precalculus which was the same as applied math except we weren't allowed to use calculators at all (well, we *could* for homework, but that's different). We learned to use TI-83's in physics, but even we didn't use them much (once in awhile). I found that my teachers were reasonably smart, and I did learn a lot in highschool.
Quote from: DarkVirus on December 21, 2003, 10:13 PM
I had asked him HOW logs worked because of course like most programmers, I want to know how people did logs BEFORE calculators and other bits of technology were around.
...
I just think he thought the students wouldn't really care about the actual method.
I think they typically looked them up in a table. You would have a thick book with numbers in which you could look up say 1.23456 and get the base 10 logarithm of it, 0.0915122. There might also have been graphs of logarithms that you could use to intepolate out your answers. A table wouldn't need to cover other numbers than say 1 <= x < 10, since log10(12.3456) = 1 + log10(1.23456).
I don't think there's any practical way of calculating logarithms, sines for angles and similar functions without using either known values, tables or infinitely repeating formulas (where you keep adding terms until you have the number of significant digits that you want).
Quote from: DarkVirus on December 21, 2003, 10:13 PM
I had asked him HOW logs worked because of course like most programmers, I want to know how people did logs BEFORE calculators and other bits of technology were around.
We used tables like Adron said. A lot of old mathematicians spend their lives writing tables. There are tables for logarithms, trigonometric functions, 1/x and much more. You can even find a table for pi and its decimals
We also used slide rules (http://search.britannica.com/search?ref=A01015&query=slide+rule&exact) for calculations.