« previous next »
Pages: 1 ... 17 18 [19] 20 21 ... 46   Go Down

Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 88606 times)

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #270 on: October 26, 2010, 03:30:42 pm »
I've tried it so many times, can't seem to get it!

Simplify...



Many thanks!
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: C1 DOUBTS HERE!!!!!
« Reply #271 on: October 26, 2010, 04:02:10 pm »
Quote from: Ivo on October 26, 2010, 03:30:42 pm
I've tried it so many times, can't seem to get it!

Simplify...



Many thanks!

Is the answer 13/3 + (40,/3)/9  ???

,/ represents square root
Logged

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #272 on: October 26, 2010, 05:04:48 pm »
Quote from: Deadly_king on October 26, 2010, 04:02:10 pm
Is the answer 13/3 + (40,/3)/9  ???

,/ represents square root

Indeed it its!  Could you please show how you got it?

I took out a factor of \sqrt{3}, but could never get the answer!
« Last Edit: October 26, 2010, 05:07:02 pm by Ivo »
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: C1 DOUBTS HERE!!!!!
« Reply #273 on: October 26, 2010, 06:18:19 pm »
Quote from: Ivo on October 26, 2010, 05:04:48 pm
Indeed it its!  Could you please show how you got it?

I took out a factor of \sqrt{3}, but could never get the answer!

OK....i'll elaborate.

I'll simplify the numbers first.

(,/3)-3 + (,/3)-2 + (,/3)-1 + (,/3)0 + (,/3)1 + (,/3)2 + (,/3)3

Now we can simplify : Anything to the power of zero is 1 and a square root to the power of 2 results in the number being squared

(,/3)-3 + 3-1 + (,/3) + 1 + (,/3) + 3 + (,/3)3

Now we add the numbers only.
1 + 3 + 1/3 + (,/3)-3 + (,/3)-1 + (,/3)1 + (,/3)3

Anything to the power of negative factor can be written as 1 over the value

Example : x-1 ----> 1/x

Hence => 13/3 + 1/(,/3)3 + 1/(,/3) + (,/3) + (,/3)3

Now you need to rationalise, that is multiply by (,/3)/(,/3) so that you obtained all the roots in the numerator. Moreover (,/3)2 = 3 ---> (,/3)3 = 3(,/3)

13/3 + (,/3)/9 + (,/3)/3 + (,/3) + 3(,/3)

Now forget about the (,/3) and add its coefficients to get the required solution ;)

13/3 + 40(,/3)/9

Hope it helps :)
« Last Edit: October 29, 2010, 05:37:15 am by Deadly_king »
Logged

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #274 on: October 27, 2010, 12:47:15 am »
Quote from: Deadly_king on October 26, 2010, 06:18:19 pm
OK....i'll elaborate.

I'll simplify the numbers first.

(,/3)-3 + (,/3)-2 + (,/3)-1 + (,/3)0 + (,/3)1 + (,/3)2 + (,/3)3

Now we can simplify : Anything to the power of zero is 1 and a square root to the power of 2 results in the number being squared

(,/3)-3 + 3-1 + (,/3) + 1 + (,/3) + 3 + (,/3)3

Now we add the numbers only.
1 + 3 + 1/3 + (,/3)-3 + (,/3)-1 + (,/3)1 + (,/3)3

Anything to the power of negative factor can be written as 1 over the value

Example : x-1 ----> 1/x

Hence => 13/3 + 1/(,/3)3 + 1/(,/3) + (,/3) + (,/3)3

Now you need to rationalise, that is multiply by (,/3)/(,/3) so that you obtained all the roots in the numerator. Moreover (,/3)2 = 3 ---> (,/3)3 = 3(,/3)

13/3 + (,/3)/9 + (,/3)/3 + (,/3) + 3(,/3)

Now forget about the (,/3) and add its coefficients to get the required solution ;)

13/3 + 40(,/3)/9

Hope it helps :)

OK thanks a lot for your detailed explanation!  I just didn't understand the last bit, I somehow got \frac{39}{9}\sqrt{3}, instead of \frac{40}{9}\sqrt{3}.  Where could I have gone wrong?
« Last Edit: October 29, 2010, 05:37:44 am by Deadly_king »
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: C1 DOUBTS HERE!!!!!
« Reply #275 on: October 29, 2010, 05:49:30 am »
Quote from: Ivo on October 27, 2010, 12:47:15 am
OK thanks a lot for your detailed explanation!  I just didn't understand the last bit, I somehow got \frac{39}{9}\sqrt{3}, instead of \frac{40}{9}\sqrt{3}.  Where could I have gone wrong?

I guess you forgot 1/9. ;)

To the extent I had reached in my explanation, you just needed to add 1/9 + 1/3 + 1 + 3 = 40/9

These are all coefficients of (,/3), so it's similar to adding x/9 + x/3 + x + 3x = 40x/9

You just need to replace x by (,/3) to obtain 40(,/3)/9 :D
Logged

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #276 on: October 29, 2010, 11:51:04 am »
Quote from: Deadly_king on October 29, 2010, 05:49:30 am
I guess you forgot 1/9. ;)

To the extent I had reached in my explanation, you just needed to add 1/9 + 1/3 + 1 + 3 = 40/9

These are all coefficients of (,/3), so it's similar to adding x/9 + x/3 + x + 3x = 40x/9

You just need to replace x by (,/3) to obtain 40(,/3)/9 :D

Thanks for your explanations!  I made a schoolboy error - I thought \frac{1}{3} = \frac{2}{9}.  Silly me!  + rep for you anyway, thanks!
« Last Edit: October 30, 2010, 02:45:39 pm by Ivo »
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: C1 DOUBTS HERE!!!!!
« Reply #277 on: October 29, 2010, 02:45:22 pm »
Quote from: Ivo on October 29, 2010, 11:51:04 am
Thanks for your explanations!  I made a schoolboy error - I thought \frac{1}{3} = \frac{2}{9}.  Silly me!  + rep for you anyway, thanks!

No problem buddy ;)

Try to be more careful next time ;D

I guess you just +rep someone, so you need to wait for 2hrs to +rep another person.

Don't worry about it ;)
Logged

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #278 on: October 29, 2010, 03:11:04 pm »
Another doubt here, I think this is harder than C1, because it is apparantly an extension question from a C1 textbook.  Anyway, here's the question:

Let p(x) = x^2 - 6x - 3 and q(x) = x^2 - 2x + 4.

The polynomial p(x) + aq(x), where a is a constant, is a perfect square.

Calculate the two possible values of a.

Very hard for me, here's my working so far.  My answer's wrong though.

I found the polynomial to be: a+1(x - \frac{a-3}{a+1})^2 - \frac{a^2-6a+9}{a+1} + 4a - 3.

Then I said that because it's a perfect square: - \frac{a^2-6a+9}{a+1} + 4a - 3 = 0, and then got the quadratic: 3a^2 + 7a - 12 = 0.  This gives imaginary solutions, but there are 2 real solutions for a.

Many thanks in advance!
« Last Edit: October 29, 2010, 04:20:02 pm by Ivo »
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Tyserius

  • Actuary Wannabe
  • SF Geek
  • ****
  • Posts: 318
  • Reputation: 383
  • Gender: Male
  • You are who you choose to become
Re: C1 DOUBTS HERE!!!!!
« Reply #279 on: October 30, 2010, 03:32:32 am »
Your mistake is that it should be (a+3)/(1+a) instead of (a-3)/(1+a) as x + (-a-3)/(1+a) = x - (a+3)/(1+a). I tried many ways to solve but I don't seem to get the 2nd value for a. The first value should be a = 3?
Logged
Take it easy and go slow and steady.

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #280 on: October 30, 2010, 12:34:42 pm »
Quote from: Tyserius on October 30, 2010, 03:32:32 am
Your mistake is that it should be (a+3)/(1+a) instead of (a-3)/(1+a) as x + (-a-3)/(1+a) = x - (a+3)/(1+a). I tried many ways to solve but I don't seem to get the 2nd value for a. The first value should be a = 3?

Thanks for your hint there Tyserius!  I've solved it now:

The polynomial is: a+1(x - \frac{a+3}{a+1})^2 - \frac{(a+3)^2}{a+1} + 4a - 3

Therefore, -\frac{(a^2+6a+9)}{a+1} + 4a - 3 = 0

So quadratic: 3a^2 - 5a - 12 = 0

Giving solutions of a: 3, -\frac{4}{3}

Many thanks once again for pointing out that error, as a result an + rep comes your way!  :)
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

Offline Tyserius

  • Actuary Wannabe
  • SF Geek
  • ****
  • Posts: 318
  • Reputation: 383
  • Gender: Male
  • You are who you choose to become
Re: C1 DOUBTS HERE!!!!!
« Reply #281 on: November 03, 2010, 01:09:02 am »
I actually tried putting a = -4/3 but it didn't seem to work.. Hmm... You sure that's the answer? a=3,-4/3? Not just a=3? D:
Logged
Take it easy and go slow and steady.

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: C1 DOUBTS HERE!!!!!
« Reply #282 on: November 11, 2010, 09:13:18 pm »
I got the same as ivo
Logged

Offline Ivo

  • SF Geek
  • ****
  • Posts: 515
  • Reputation: 2767
Re: C1 DOUBTS HERE!!!!!
« Reply #283 on: November 13, 2010, 12:38:39 pm »
Here's another doubt:

Find the coordinates of the stationary points on the curve with equation y = x(x-1)^2. 
Find the set of real values of k such that the equation x(x-1)^2 = k^2 has exactly one real root.

For the 1st part, I got the stationary points as: (\frac{1}{3},\frac{4}{27}) is a maximum, (1,0) is a minimum.  I know these are right, but I'm not sure how to solve the 2nd part.

Many thanks!
Logged
Always willing to help!  8)
"In helping others, we shall help ourselves, for whatever good we give out completes the circle and comes back to us."

elemis

  • Guest
Re: C1 DOUBTS HERE!!!!!
« Reply #284 on: November 13, 2010, 01:17:27 pm »
One moment.
Logged
Pages: 1 ... 17 18 [19] 20 21 ... 46   Go Up
« previous next »