Lstigant
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#141
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#141
It makes 80(9y^2+6y^3+9y^4)
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Amy x
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#142
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#142
Name:  area.png
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Size:  5.7 KBThats what i did
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JASApplications
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#143
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#143
log2(x-3) - log2x = 4
log2(x-3)/x = 4
(x-3)/x = 2^4
(x-3)/x = 16
(x-3) = 16x
-1/5 = x

Reckon I'd get a few method marks even though the final answer is most likely wrong?
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stirkee
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#144
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#144
(Original post by olivers16)
Can someone please explian how on Q9 i) the 4/x^2 integrated and put the 2a and a into becomes 2? Thanks.
I'll try my best to help!

Basically:

You need to integrate between 2a and a: (2x3 -5x2 - 4)x-2

Which equals 2x - 5 - 4x-2

Integrating it makes it become [x2 -5x + 4x-1] between 2a and a

So therefore (4a2 - 10a + 4/2a) - (a2 - 5a + 4/a)

does that help?
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Lstigant
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#145
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#145
For 6:
just sub in values to U1 U2 and U3 to get x=15/2
put x in to find 8 works
ar/a=ar^2/ar=r
So (x+4)/2x=(2x-7)/(x+4)
goes to x^2+8x+16=4x^2-14x
3x^2-22x-16=(3x+2)(x-8)
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Lstigant
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#146
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#146
9 has been answered by someone else
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cleverslacker
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#147
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#147
I could really do with someone going through the very last parts of 1 and 9, for both of those I used the quadratic formula and ended up with two answers.
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stirkee
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#148
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(Original post by cleverslacker)
I could really do with someone going through the very last parts of 1 and 9, for both of those I used the quadratic formula and ended up with two answers.
a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic

You should get two answers from the quadratic formula for 9 ii) but one of them was negative - remember in the question it says that a is positive.
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olivers16
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#149
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(Original post by stirkee)
I'll try my best to help!

Basically:

You need to integrate between 2a and a: (2x3 -5x2 - 4)x-2

Which equals 2x - 5 - 4x-2

Integrating it makes it become [x2 -5x + 4x-1] between 2a and a

So therefore (4a2 - 10a + 4/2a) - (a2 - 5a + 4/a)

does that help?
Ahh, yes it does Thank you. I found I got the first two values but not the +2
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Lstigant
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#150
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(Original post by stirkee)
a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic
Makes more sense to use completing the square as they want exact answer and is quicker tbh, only (1-sqrt7)/3=a as it said a was a positive constant
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mossss
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#151
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#151
what did everyone get for 6iib) ?
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cleverslacker
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#152
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(Original post by stirkee)
a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic
thanks but it was when trying to find the roots of the quadratic that I had the problem. it says "other possible value" not "values", but I got two: (1/3)+(1/3)sqrt(7), (1/3)-(1/3)sqrt(7)
Anyone else get this?
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stirkee
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#153
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#153
(Original post by cleverslacker)
thanks but it was when trying to find the roots of the quadratic that I had the problem. it says "other possible value" not "values", but I got two: (1/3)+(1/3)sqrt(7), (1/3)-(1/3)sqrt(7)
Anyone else get this?
See my edit

it said in the question that a is a positive constant.
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Selvey
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#154
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#154
(Original post by mossss)
what did everyone get for 6iib) ?
My earlier post:

(Original post by Selvey)


Not sure if the below 'x' values are right (I can't remember them exactly), but this is the method I used.

a=2x
ar = x+4
ar2= 2x-7

Therefore, as 'ar
2'÷'ar' = r, that can be set equal to 'ar'÷'a'.Therefore (2x-7)÷(x+4)=(x+4)÷(2x). From there I cross multiplied and solved; and as one of the solutions was 8, I knew the other one was correct.(If I remember correctly x was -2/3).
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cleverslacker
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#155
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#155
(Original post by Lstigant)
Makes more sense to use completing the square as they want exact answer and is quicker tbh, only (1-sqrt7)/3=a as it said a was a positive constant
Oh bugger I didn't realise it said "positive" constant. Dya guys think I'll get like 5/6 marks?
Thanks for all your help in this thread
btw. I think you mean a=(1+sqrt7)/3, that would be the positive one
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Lstigant
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#156
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(Original post by mossss)
what did everyone get for 6iib) ?
ar/a=ar^2/ar=r
So (x+4)/2x=(2x-7)/(x+4)
goes to x^2+8x+16=4x^2-14x
3x^2-22x-16=(3x+2)(x-8)
x=8 or x=-2/3
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stirkee
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#157
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#157
(Original post by cleverslacker)
Oh bugger I didn't realise it said "positive" constant. Dya guys think I'll get like 5/6 marks?
I'd imagine that putting both roots from the quadratic would only lose you 1 mark yeah, with the rest of the working being correct!
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cleverslacker
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#158
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#158
(Original post by stirkee)
See my edit

it said in the question that a is a positive constant.
(Original post by stirkee)
I'd imagine that putting both roots from the quadratic would only lose you 1 mark yeah, with the rest of the working being correct!
Cool, thanks.
Not looking tooo bad for me, will have to check Mr. M's for silly mistakes though.
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stirkee
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#159
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#159
(Original post by cleverslacker)
Cool, thanks.
Not looking tooo bad for me, will have to check Mr. M's for silly mistakes though.
Hopefully you do well!

I know I've lost 2 marks (maybe 3 depending on how many working marks I get) so hopefully I've made few/no stupid mistakes..

From what I've seen in this thread, a lot of the answers that I got (and can remember..) are being repeated so fingers crossed!!
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Namrah
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#160
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#160
(Original post by KRJACK)
What did anyone get for the value of k?
2/3 rings a bell? Or it might have been 3/2? Can't remember if I got +/- either?
Sorry, im not very helpful
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