How do you solve #x^2 - x - 1 = 0#?

Question

1186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-04T11:50:53

Use the quadratic formula or complete the square to find:

#x = (1+-sqrt(5))/2#

Method 1 - Quadratic Formula

#x^2-x-1# is of the form #ax^2+bx+c#, with #a=1#, #b=-1# and #c=-1#.

This has solutions given by the quadratic formula

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-1)+-sqrt((-1)^2 - (4xx1xx(-1))))/(2*1)#

#=(1+-sqrt(1+4))/2 = (1+-sqrt(5))/2#

#0 = x^2-x-1 = (x-1/2)^2-(1/2)^2-1#

#= (x-1/2)^2-5/4#

Add #5/4# to both ends to get:

#(x-1/2)^2 = 5/4#

So:

#x-1/2 = +-sqrt(5/4) = +-sqrt(5)/sqrt(4) = +-sqrt(5)/2#

Add #1/2# to both sides to get:

#x = 1/2+-sqrt(5)/2 = (1+-sqrt(5))/2#

How are the two methods related?

#0 = ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#

Hence:

#(x+b/(2a))^2 = (b^2/(4a)-c)/a = (b^2 - 4ac)/(4a^2)#

So:

#x+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)#

#=+-sqrt(b^2 - 4ac)/(2a)#

Subtract #b/(2a)# from both sides to get:

#x = (-b+-sqrt(b^2-4ac))/(2a)#