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The following solution was done by Ling Xiang Ning, Raffles Institution, Singapore.
Using the quadratic formula to solve the equation
$ x^2 = 7x + 1$
$x^2 - 7x - 1 = 0$
I find that $x$ is $(7 \pm \sqrt{53} )/2$. The positive solution is approximately $7.140054945$
This equation is equivalent to $x = 7 + 1/x$ and hence to the sequence of continued fractions mentioned in the problem. These continued fractions give better and better approximations to the positive root of the quadratic equation and I shall do them one by one.
$ 7+\frac{1}{7} = \frac{50}{7} = 7.142857142 $
$ 7+\frac{1}{7+\frac{1}{7}} = \frac{357}{50} = 7.14 $
$ 7+\frac{1}{7+\frac{1}{7+\frac{1}{7}}} = \frac{2549}{357} = 7.140056022 $
$7+\frac{1}{7+\frac{1}{7+\frac{1}{7 + \frac{1}{7}}}} = \frac{18200}{2549} = 7.140054923 $
To find a rational approximation to $\sqrt{53}$ we take, as above, \[ \frac{7 + \sqrt{53}}{2} \approx \frac{2549}{357} \]
which gives \[ \sqrt{53} \approx 2 (\frac{2549}{357}) - 7 \approx \frac{2599}{357}. \]
$ 5+\frac{1}{5} = \frac{26}{5} = 5.2 $
$5+\frac{1}{5+\frac{1}{5}} = \frac{135}{26} = 5.192307692 $
$5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}} = \frac{701}{135} = 5.192592592 $
$5+\frac{1}{5+\frac{1}{5+\frac{1}{5+\frac{1}{5}}}} = \frac{3640}{701} = 5.192582025 $
Similarly, using the equation, $x^2 = 5x + 1$, which has solutions $\frac{5\pm \sqrt{29} }{2}$ , we can find a rational approximation to $\sqrt{29}$. The positive root is approximately $5.192582404$. The sequence of continued fractions is:
As you can see, the sequence of continued fractions gives better and better approximations to the positive root of the quadratic equation. \par Using $\frac{5\pm \sqrt{29}}{2} \approx \frac{3640}{701}$ gives $\frac{3775}{701}$ as a rational approximation to $\sqrt{29}$.Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?