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Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 98474 times)

Offline Ivo

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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!
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Offline Deadly_king

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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
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Offline Ivo

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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 »
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Offline Deadly_king

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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 »
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Offline Ivo

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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 »
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Offline Deadly_king

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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
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Offline Ivo

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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 »
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Offline Deadly_king

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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 ;)
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Offline Ivo

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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 »
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Offline Tyserius

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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?
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Offline Ivo

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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!  :)
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Offline Tyserius

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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:
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Offline astarmathsandphysics

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Re: C1 DOUBTS HERE!!!!!
« Reply #282 on: November 11, 2010, 09:13:18 pm »
I got the same as ivo
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Offline Ivo

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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!
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Re: C1 DOUBTS HERE!!!!!
« Reply #284 on: November 13, 2010, 01:17:27 pm »
One moment.
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