# How do you solve using the completing the square method t^2 - t - 1 = 0?

Jan 23, 2017

$t = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = 2 t - 1$ and $b = \sqrt{5}$

We can premultiply the given equation by $4$ to cut down on the arithmetic involving fractions:

$0 = 4 \left({t}^{2} - t - 1\right)$

$\textcolor{w h i t e}{0} = 4 {t}^{2} - 4 t - 4$

$\textcolor{w h i t e}{0} = 4 {t}^{2} - 4 t + 1 - 5$

$\textcolor{w h i t e}{0} = {\left(2 t - 1\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 t - 1\right) - \sqrt{5}\right) \left(\left(2 t - 1\right) + \sqrt{5}\right)$

$\textcolor{w h i t e}{0} = \left(2 t - 1 - \sqrt{5}\right) \left(2 t - 1 + \sqrt{5}\right)$

Hence:

$t = \frac{1}{2} \left(1 \pm \sqrt{5}\right) = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

$\textcolor{w h i t e}{}$
Footnote

These solutions of ${t}^{2} - t - 1 = 0$ are especially useful in relation to the Fibonacci sequence:

$0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , \ldots$

where each term is the sum of the two previous ones.

${F}_{0} = 0$

${F}_{1} = 1$

${F}_{n + 2} = {F}_{n + 1} + {F}_{n}$

The further you go in the Fibonacci sequence, the closer the ratio between successive terms gets to $\frac{1}{2} + \frac{\sqrt{5}}{2} \approx 1.618034$, known as the golden ratio, sometimes given the symbol $\varphi$.

In fact the general term of the Fibonacci sequence is given by the formula:

${F}_{n} = \frac{{\varphi}^{n} - {\left(- \varphi\right)}^{- n}}{\sqrt{5}}$

The golden ratio has many interesting properties, particularly in relation to regular pentagons, dodecahedrons and icosahedrons.

It is also the slowest converging continued fraction:

varphi = [1;bar(1)] = 1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+...))))))))