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

##### 1 Answer

Use the quadratic formula or complete the square to find:

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

#### Explanation:

**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#

**Method 2 - Completing the square**

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

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

Add

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

So:

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

Add

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

**How are the two methods related?**

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

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