First slide

Theory of equations

Question

$The value of 3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + \frac{1}{3 + \dots \infty}}} is equal to}$

Moderate

A

$4 + \sqrt{3}$
B

$2 + \sqrt{3}$
C

$3 + 2 \sqrt{3}$
D

$1.5 + \sqrt{3}$

Solution

The given expression can be written as
$\begin{matrix} x = 3 + \frac{1}{4 + \frac{1}{x}} \\ = 3 + \frac{x}{4 x + 1} \\ x - 3 = \frac{x}{4 x + 1} \\ (x - 3) (4 x + 1) = x \\ 4 x^{2} + x - 12 x - 3 = x \\ 4 x^{2} - 12 x - 3 = 0 \\ x = \frac{12 \pm \sqrt{144 + 48}}{8} \\ = \frac{12 \pm \sqrt{192}}{8} \\ = \frac{3}{2} \pm \sqrt{3} \\ = 1.5 + \sqrt{3} \end{matrix}$

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