Finding the quadratic polynomial of a continued fraction

Question

Given $\alpha=[3,\bar{2},\bar{4},\bar{5}]=3+\tfrac{1}{2+\tfrac{1}{4+\tfrac{1}{5+...}}}$

I want to (i) find the quadratic integer polynomial P for which $\alpha$ is a root and (ii) hence write $\alpha$ as a quadratic irrational.

Here is what I did , could anyone tell me if this is correct:

i)

$\alpha=3+\tfrac{1}{2+\tfrac{1}{4+\tfrac{1}{5+...}}}=3+\tfrac{1}{2+\tfrac{1}{4+\tfrac{1}{5-3+\alpha}}}=3+\tfrac{1}{2+\tfrac{1}{4+\tfrac{1}{2+\alpha}}}=\tfrac{1}{2+\tfrac{1}{\tfrac{9+4\alpha}{2+\alpha}}}=3+\tfrac{1}{\tfrac{20+9\alpha}{9+4\alpha}}=\tfrac{69+22\alpha}{20+9\alpha}$

$\therefore$ $\alpha(20+9\alpha)=69+22\alpha$ which means the quadratic integer polynomial is $9\alpha^2-2\alpha-69=0$

ii) Can we just use the quadratic formula ? Then $\alpha=\tfrac{2^+_-\sqrt{4-2484}}{18}$

Answer 1

not sure where you went wrong, it is $$ \alpha = \frac{11+\sqrt{2605}}{18} $$ and $$ 9 \alpha^2 - 11 \alpha - 69 = 0 $$

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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 9 -11 -69

  0  form              9         -11         -69  delta      0
  1  form            -69          11           9  delta      3
  2  form              9          43         -21


          -1          -3
           0          -1

To Return  
          -1           3
           0          -1

0  form   9 43 -21   delta  -2
1  form   -21 41 11   delta  4
2  form   11 47 -9   delta  -5
3  form   -9 43 21   delta  2          -1 composed with form zero  
4  form   21 41 -11   delta  -4
5  form   -11 47 9   delta  5
6  form   9 43 -21


  form   9 x^2  + 43 x y  -21 y^2 

minimum was   9rep   x = 1   y = 0 disc 2605 dSqrt 51  M_Ratio  32.11111
Automorph, written on right of Gram matrix:  
-205  -1071
-459  -2398
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$