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

1 Answer
Jan 23, 2017

Answer:

#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+...))))))))#