Just wondering, what happened to your 2a in the quadratic formula? I had a (1+i) in the bottom of the fraction, which would explain why they asked "express your answer in its simplest form", i.e multiplying through by the complex conjugate (1i)/(1i), I checked my solution with a ex Cambridge Maths guy and we both seemed to get the same bit. so your answer / (1+i)(Original post by DFranklin)
Step QI, P5.
iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse..2z=2\pm\sqrt{4  4(1+i)(2\sqrt{3}2)} \\
\implies z = 1 \pm \sqrt{1  (1+i)(2\sqrt{3}2)} = 1 \pm \sqrt{32\sqrt{3}+2(1\sqrt{3})i)
Using (ii),
.
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 16062010 11:16

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 16062010 11:20
Quite possible; I seem to recall losing the will to live with that question about half the way through...

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 123
 17062010 12:10
What do you think the grade boundary for a 1 would be for this paper?

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 20062010 21:45
Shall we add the amendment to that then? Funny you mention that, This was a question which I got the knack of quite quickly (as opposed to many others which took me literally years)

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 125
 21062011 16:53
(Original post by insparato)
Typing up STEP III question 2 Now
 eqn 1
 eqn 2
 eqn 3
eqn (2  1)
eqn4
sub eqn 4 into 3
eqn 5
sub eqn 4 into 2
eqn6
subtracting 6 from 5
Sub into eqn 5
Sub these back into eqn 1 just to see they are indeed satisfy the system.
Therefore
Not complete, the cases A = 0, A = 1 have to be investigated as the standard solution blows up when A = 0 and A = 1. Ive done alot of maths today so i wont be doing this tonight but i might have a go tomorrow or something... Anyone who feels up to finishing it can do so freely .
If anyone requests the solution i can always do a copy and scan in just message me 
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 126
 26072011 14:49
Here is an alternative approach to STEP 1996 I Q10.

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 127
 31052012 12:55
(Original post by themaths)
Such a nice alternative solution with matricies i would type it out but i do ot know how to latex a matrix o.o sorry guys...
If anyone requests the solution i can always do a copy and scan in just message me
Last edited by brianeverit; 31052012 at 12:59. 
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 128
 06082012 15:42
(Original post by DFranklin)
Step QI, P5.
(i) . Suppose . Then
would give us a rational expression for , contradiction.
So .
As r is rational, . (i.e. our solutions are .
(ii) . So
. Again, we set up the quadratic and get:
. Use the quadratic formula to get:
Now .
So .
As in (i), we need to find the square root of this, so look for rational with .
We find and end up with a quadratic . Solve to find and since we want rational deduce and so the square root is .
Plugging this back in, we have
.
Since p is real, we take the latter root, dividing by 2 and using (i) we find . Use to get final solutions .
(iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse..2z=2\pm\sqrt{4  4(1+i)(2\sqrt{3}2)} \\
\implies z = 1 \pm \sqrt{1  (1+i)(2\sqrt{3}2)} = 1 \pm \sqrt{32\sqrt{3}+2(1\sqrt{3})i)
Using (ii),
.Last edited by brittanna; 06082012 at 15:44. 
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 30122012 14:48
(Original post by DFranklin)
Step II, Q3:
F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.
F0.F2F1^2 = 1, F1.F3F2^2 = 1, F2.F4F3^2 = 1.
Claim . Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then
So true for n=k+1 and so true for all n by induction.
Want to show . Assume true for 1 < k <=m. Then
.
So true for k=m+1.
Explictly when k=1 we have . So true for k=1, so true for all k.
For the second part, you have assumed that takes the same form as isn't that defeating the whole point of the induction?
We assume n=k, but in the proof it seems n=k1 is assumed too, am I talking a load of rubbish? 
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 130
 30122012 15:17
The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

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 31122012 12:03
(Original post by BabyMaths)
The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

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 132
 04022013 19:35
(Original post by DFranklin)
Step QI, P5.
(i) . Suppose . Then
would give us a rational expression for , contradiction.
So .
As r is rational, . (i.e. our solutions are .
(ii) . So
. Again, we set up the quadratic and get:
. Use the quadratic formula to get:
Now .
So .
As in (i), we need to find the square root of this, so look for rational with .
We find and end up with a quadratic . Solve to find and since we want rational deduce and so the square root is .
Plugging this back in, we have
.
Since p is real, we take the latter root, dividing by 2 and using (i) we find . Use to get final solutions .
(iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse..2z=2\pm\sqrt{4  4(1+i)(2\sqrt{3}2)} \\
\implies z = 1 \pm \sqrt{1  (1+i)(2\sqrt{3}2)} = 1 \pm \sqrt{32\sqrt{3}+2(1\sqrt{3})i)
Using (ii),
. 
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 133
 05032013 19:26
For III Q2.
You can show that those are the answers are correct with the exception of a=1,0. Here's how:
The three equation can be rewritten as three matrices as shown below:
We can now solve it because of the result
Ainverse (A)R = Ainverse B
which goes to :
R = Ainverse B
now this means that the equations have no solution when A is noninvertible.
A is invertible when
so when
so when orLast edited by Rainingshame; 12042013 at 16:24. 
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 134
 01042013 00:29
(Original post by DFranklin)
Quite possible; I seem to recall losing the will to live with that question about half the way through...
Q5 STEP I:
Part i)
Part ii)
Part iii)

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 135
 10042013 12:20
(Original post by DFranklin)
Solved it, yes, written it up, no  nice job. I didn't see: did you manage the last bit? I thought it surprisingly tricky; needing to find the value of tan(3pi/8) was a tad annoying...
Notice that you're letting a=cos(3pi/4), use t substitutions with t=tan(3pi/8) and it all cancels down very quickly to 2arctan(1) 
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 136
 11042013 13:12
(Original post by DFranklin)
Step II, Q3:
F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.
F0.F2F1^2 = 1, F1.F3F2^2 = 1, F2.F4F3^2 = 1.
Claim . Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then
So true for n=k+1 and so true for all n by induction.
Want to show . Assume true for 1 < k <=m. Then
.
So true for k=m+1.
Explictly when k=1 we have . So true for k=1, so true for all k. 
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 137
 11042013 13:39
(Original post by Nick_)
.. 
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 11042013 17:04
(Original post by DFranklin)
I explicitly show the result for n<=3 in the first few lines of the proof. 
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 139
 11042013 17:32
Fair point.

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 140
 12042013 15:44
(Original post by Rainingshame)
For III Q2.
You can show that those are the answers are correct with the exception of a=1,0. Here's how:
The three equation can be rewritten as three matrices as shown below:
We can now solve it because of the result
Ainverse (A)R = Ainverse B
which goes to :
R = Ainverse B
now this means that the equations have no solution when A is invertible.
A is invertible when
so when
so when or
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Updated: May 25, 2015
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