Author Topic: For anyone who is good at math... (Read 6976 times)

tjanuranus · « **Reply #35 on:** February 26, 2011, 12:40:19 PM »

omg so easy once i read your post. I thought i could do it the way i was doing it. Thanks!

Jamesman42 · « **Reply #36 on:** February 26, 2011, 05:42:22 PM »

Quote from: Fiery Winds on February 26, 2011, 12:25:52 PM

You can't take the square root of two terms added or subtracted together. Move the -b^2 term to the other side first and you'll get your answer. You got lucky with your first answer.

DAT MATHEMATIXX POWAH SON. WE OWN THE PLACE

tjanuranus · « **Reply #37 on:** February 26, 2011, 05:52:57 PM »

got my first test next saturday so i'm studying up and i was so pissed off when i couldn't get that correct answer because it's one of the simple things on the dest. Actually the harder stuff i've been doing better at. lol. Well for some reason that question stumped me but now i see it's easy.

tjanuranus · « **Reply #38 on:** March 03, 2011, 11:50:38 AM »

ok so it turns out my book is wrong quite a lot. This book sucks! i'm practicing for my test and the book has the answers for the odd practice questions. But sometimes the answers aren't correct. I'm trying to see if i got this right or not because the book says i'm wrong..

x^3 + 3 = 3x^2 + x

so i did...

x^3 - 3x^2 - x + 3 = 0 Step 1

x^2(x-3) -(x+3) = 0 Step 2

x^2 = 0 x=0
x-3 = 0 x=3
-x-3 = 0 x=-3

So my answer is x=0, +-3 But the book says x=+-1, +-3

Orbert · « **Reply #39 on:** March 03, 2011, 11:59:27 AM »

You're not done after Step 2. You can't just factor from there.

tjanuranus · « **Reply #40 on:** March 03, 2011, 12:08:39 PM »

then i have no idea what to do because the other example in the book factors from that point. Confused!

tjanuranus · « **Reply #41 on:** March 03, 2011, 12:12:44 PM »

Got it now. Thanks

Fiery Winds · « **Reply #42 on:** March 03, 2011, 12:43:10 PM »

Quote from: tjanuranus on March 03, 2011, 11:50:38 AM

ok so it turns out my book is wrong quite a lot. This book sucks! i'm practicing for my test and the book has the answers for the odd practice questions. But sometimes the answers aren't correct. I'm trying to see if i got this right or not because the book says i'm wrong..

x^3 + 3 = 3x^2 + x

so i did...

x^3 - 3x^2 - x + 3 = 0 Step 1

x^2(x-3) -(x+3) = 0 Step 2

x^2 = 0 x=0
x-3 = 0 x=3
-x-3 = 0 x=-3

So my answer is x=0, +-3 But the book says x=+-1, +-3

Just making sure you understand the error, the bolded part is incorrect. It should be (x-3) due to the distributive property of the negative sign. That then allows you to factor and get the following:

(x^2 - 1)(x-3) = 0 with 3 solutions: x= +-1, 3 (not -3). It's a 3rd order polynomial, so you'll have at most 3 solutions.

tjanuranus · « **Reply #43 on:** April 11, 2011, 03:34:13 AM »

Ok my mind is blanking on this. Either there is NO way to do this or i'm having a brain freeze. Could use any help. How do you do this?

(3x-1) / (-4x+6)

Also is this correct?
(3x-1) * (-4x+6)

-12x^2 + 18x + 4x -6
-12x^2 + 22x - 6

Answer..
-2(6x^2 = 11x + 3)

Thanks!

Jamesman42 · « **Reply #44 on:** April 11, 2011, 04:53:31 AM »

Quote from: tjanuranus on April 11, 2011, 03:34:13 AM

Ok my mind is blanking on this. Either there is NO way to do this or i'm having a brain freeze. Could use any help. How do you do this?

(3x-1) / (-4x+6)

Hard to tell without any directions given whatsoever.

Quote from: tjanuranus on April 11, 2011, 03:34:13 AM

Also is this correct?
(3x-1) * (-4x+6)

-12x^2 + 18x + 4x -6
-12x^2 + 22x - 6

Answer..
-2(6x^2 = 11x + 3)

Thanks!

Here is the thing that is confusing me: Every step has no equal sign, and then all of a sudden the answer has an equal sign. You can't do that, I am 99% sure you are not including all the information needed or the directions given (or both).

tjanuranus · « **Reply #45 on:** April 11, 2011, 04:56:12 AM »

Quote from: Jamesman on April 11, 2011, 04:53:31 AM

Quote from: tjanuranus on April 11, 2011, 03:34:13 AM
Ok my mind is blanking on this. Either there is NO way to do this or i'm having a brain freeze. Could use any help. How do you do this?

(3x-1) / (-4x+6)

Hard to tell without any directions given whatsoever.

Quote from: tjanuranus on April 11, 2011, 03:34:13 AM
Also is this correct?
(3x-1) * (-4x+6)

-12x^2 + 18x + 4x -6
-12x^2 + 22x - 6

Answer..
-2(6x^2 = 11x + 3)

Thanks!

Here is the thing that is confusing me: Every step has no equal sign, and then all of a sudden the answer has an equal sign. You can't do that, I am 99% sure you are not including all the information needed or the directions given (or both).

that was a typo it's supposed to be a - sign.

And the questions are...

f(x) = 3x-1
g(x) = -4x + 6

question 1..

(fg)(x)= ?

question 2

(f/g)(x)= ?

73109 · « **Reply #46 on:** April 11, 2011, 05:50:21 AM »

I love those.

Jamesman42 · « **Reply #47 on:** April 11, 2011, 06:06:22 AM »

(fg)(x) = f(x)*g(x) = (3x-1)(-4x+6) = 2(3x-1)(-2x+3)

(f/g)(x) = f(x)/g(x) = (3x-1)/(-4x+6) = (3x-1)/[2(-2x+3)], x ≠ 3/2

If I am reading that notation correctly

tjanuranus · « **Reply #48 on:** April 11, 2011, 06:49:22 AM »

Quote from: Jamesman on April 11, 2011, 06:06:22 AM

(fg)(x) = f(x)*g(x) = (3x-1)(-4x+6) = 2(3x-1)(-2x+3)

(f/g)(x) = f(x)/g(x) = (3x-1)/(-4x+6) = (3x-1)/[2(-2x+3)], x ≠ 3/2

If I am reading that notation correctly

Can you go over how you came up with the first one? Why is mine wrong?

Jamesman42 · « **Reply #49 on:** April 11, 2011, 06:59:06 AM »

You got this: -2(6x^2 = 11x + 3)

Somehow, you inserted an equal sign into the expression. Expressions don't have equal signs; only equations do (note that the word equation contains the word "equal", pretty much anyway).

Also, I could FOIL (3x-1)(-4x+6) to get -12x^2 +22x - 6. What it seems like you did is you set it equal to zero, which would insert an equal sign into the expression, making it an equation. But (fg)(x) implies that they just want you to find the new form of the function, which means no equations....just expressions. So, no setting equal to 0.

I am tired, so if any of that didn't make sense, let me know!

tjanuranus · « **Reply #50 on:** April 11, 2011, 07:00:16 AM »

Quote from: Jamesman on April 11, 2011, 06:59:06 AM

You got this: -2(6x^2 = 11x + 3)

Somehow, you inserted an equal sign into the expression. Expressions don't have equal signs; only equations do (note that the word equation contains the word "equal", pretty much anyway).

Also, I could FOIL (3x-1)(-4x+6) to get -12x^2 +22x - 6. What it seems like you did is you set it equal to zero, which would insert an equal sign into the expression, making it an equation. But (fg)(x) implies that they just want you to find the new form of the function, which means no equations....just expressions. So, no setting equal to 0.

I am tired, so if any of that didn't make sense, let me know!

equal sign was a TYPO! supposed to be a Minus sign. So was my answer wrong? thanks!

Jamesman42 · « **Reply #51 on:** April 11, 2011, 07:02:13 AM »

You got it right, sorry, missed that it was a typo!

tjanuranus · « **Reply #52 on:** April 11, 2011, 03:05:46 PM »

is the division answer you wrote the only possible answer?

Fiery Winds · « **Reply #53 on:** April 11, 2011, 04:30:31 PM »

It depends on what answer your book wants. His answer and your answers are equivalent, you'll need to know whether the book wants you to FOIL it out completely or leave them as binomial factors.

Jamesman42 · « **Reply #54 on:** April 12, 2011, 01:00:28 AM »

Quote from: Fiery Winds on April 11, 2011, 04:30:31 PM

It depends on what answer your book wants. His answer and your answers are equivalent, you'll need to know whether the book wants you to FOIL it out completely or leave them as binomial factors.

For the division one?

Fiery Winds · « **Reply #55 on:** April 12, 2011, 01:07:05 AM »

Quote from: Jamesman on April 12, 2011, 01:00:28 AM

Quote from: Fiery Winds on April 11, 2011, 04:30:31 PM
It depends on what answer your book wants. His answer and your answers are equivalent, you'll need to know whether the book wants you to FOIL it out completely or leave them as binomial factors.

For the division one?

Yes.

Jamesman42 · « **Reply #56 on:** April 12, 2011, 01:15:03 AM »

I'm not following

Fiery Winds · « **Reply #57 on:** April 12, 2011, 01:26:44 AM »

Ok, I sort of mixed up his multiplication problem with the division, but the point still stands, lol. His book may want him to keep them factored like you have it:

(f/g)(x) = (3x-1)/[2(-2x+3)]

Or in simplified form:

(f/g)(x) = (3x-1)/(-4x+6)

Both are correct, but the first is "neater" than the other.

Jamesman42 · « **Reply #58 on:** April 12, 2011, 01:34:54 AM »

Oh, true haha....you had me worried that I was missing something obvious

I was also thinking an even neater answer would be

(f/g)(x) = (3x-1)/[2(3-2x)]

But that is just being picky about using less symbols. I still teach that to kids, though, because I see it happen too often where they don't keep their work neat/organized, and it trips them up. But that is another thread

Fiery Winds · « **Reply #59 on:** April 12, 2011, 01:37:57 AM »

Yeah, that's a tough situation for me. Do I get rid of the leading negative sign or put the variable first? Hmmmmmm......

Jamesman42 · « **Reply #60 on:** April 12, 2011, 01:39:10 AM »

Yeah, I have spent some good time devoted to thinking about that lol

kári · « **Reply #61 on:** April 12, 2011, 02:41:14 AM »

EDIT: Sorry forgot to scroll down to the last post.

ThroughHerEyesDude6 · « **Reply #62 on:** April 12, 2011, 03:03:58 AM »

This is gonna sound silly, but I truly enjoyed reading this. I haven't had a math class since high school. My AP Calculus score bypassed college math. So, reading this gave a me a great feeling, because I understood everything you all were talking about.

<3

tjanuranus · « **Reply #63 on:** April 12, 2011, 05:54:37 AM »

Ok I just want to make sure i am doing this right because the answer looks wrong but i checked it and it works out. Here is the question...

g(x) = 10
....
x-1

what's the inverse of that? My answer is

10
.... + 1
x

Here is how i came up with that and the checks..

photo by tjanuranus, on Flickr

Jamesman42 · « **Reply #64 on:** April 12, 2011, 06:28:00 AM »

You got it

tjanuranus · « **Reply #65 on:** April 12, 2011, 06:37:54 AM »

wow awesome! The answer looked so similar to the question that i thought maybe i was wrong but the math seemed to work out.

Jamesman42 · « **Reply #66 on:** April 12, 2011, 06:39:27 AM »

Haha yes...plus, you obviously double-checked it, and you should get y = x for both checks.

tjanuranus · « **Reply #67 on:** April 29, 2011, 10:58:48 PM »

Ok got a test in 7 and a half hours and there is a problem which for some reason i can't see. Either the book is wrong or i'm just blind.

(x^2 - 5x + 4) / (x^2-1)

can someone explain how the vertical asymptote is x = -1 and there is a hole at x =1?

because i get a vertical at +- 1 and no hole.

mrjazzguitar · « **Reply #68 on:** April 29, 2011, 11:07:21 PM »

Quote from: Jamesman on February 23, 2011, 06:11:02 PM

First, x = -1 and 4. And these are important. Double check your factoring.

Draw a number line. Label your x-values (-1, 4). Also, label an arbitrary point between the two numbers, and a point on the opposite sides of both numbers. For example, I labeled -2, 0, and 10. You are going to test all the intervals, from (-inf, -1), (-1,4), and (4, inf).

For x = -2 in the original equation, you get a positive number. Therefore, on (-inf,-1), that interval is positive. The same is true for x = 10, so (4, inf) is an interval on which the polynomial is positive.

For x = 0, it becomes negative. Therefore, the interval (-1, 4) gives you a negative number for that polynomial.

Testing a point on an interval takes care of that whole interval.

Hope that helps!

exactly.

yea what's a polynomial

tjanuranus · « **Reply #69 on:** April 29, 2011, 11:09:46 PM »

Nm got it now!!!

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Author Topic: For anyone who is good at math... (Read 6976 times)