How do you solve using the completing the square method t^2 - t - 1 = 0t2−t−1=0?
1 Answer
Explanation:
The difference of squares identity can be written:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
We will use this with
We can premultiply the given equation by
0 = 4(t^2-t-1)0=4(t2−t−1)
color(white)(0) = 4t^2-4t-40=4t2−4t−4
color(white)(0) = 4t^2-4t+1-50=4t2−4t+1−5
color(white)(0) = (2t-1)^2-(sqrt(5))^20=(2t−1)2−(√5)2
color(white)(0) = ((2t-1)-sqrt(5))((2t-1)+sqrt(5))0=((2t−1)−√5)((2t−1)+√5)
color(white)(0) = (2t-1-sqrt(5))(2t-1+sqrt(5))0=(2t−1−√5)(2t−1+√5)
Hence:
t = 1/2(1+-sqrt(5)) = 1/2+-sqrt(5)/2t=12(1±√5)=12±√52
Footnote
These solutions of
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
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+...))))))))
