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),
.
STEP Maths I, II, III 1996 Solutions
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Quite possible; I seem to recall losing the will to live with that question about half the way through...

What do you think the grade boundary for a 1 would be for this paper?

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)

(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 
Here is an alternative approach to STEP 1996 I Q10.

(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

(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),
. 
(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? 
The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

(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.

(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),
. 
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 or 
(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)

(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) 
(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. 
(Original post by Nick_)
.. 
(Original post by DFranklin)
I explicitly show the result for n<=3 in the first few lines of the proof. 
Fair point.

(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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