Question

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

Answer 1

`t = 1/2+-sqrt(5)/2`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We will use this with `a=2t-1` and `b=sqrt(5)`

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

`0 = 4(t^2-t-1)`

`color(white)(0) = 4t^2-4t-4`

`color(white)(0) = 4t^2-4t+1-5`

`color(white)(0) = (2t-1)^2-(sqrt(5))^2`

`color(white)(0) = ((2t-1)-sqrt(5))((2t-1)+sqrt(5))`

`color(white)(0) = (2t-1-sqrt(5))(2t-1+sqrt(5))`

Hence:

`t = 1/2(1+-sqrt(5)) = 1/2+-sqrt(5)/2`

`color(white)()`
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,...`

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 `1/2+sqrt(5)/2 ~~ 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 = (varphi^n-(-varphi)^(-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+...))))))))`