How do you solve x^2 - x - 1 = 0x2x1=0?

1 Answer
Oct 4, 2015

Use the quadratic formula or complete the square to find:

x = (1+-sqrt(5))/2x=1±52

Explanation:

Method 1 - Quadratic Formula

x^2-x-1x2x1 is of the form ax^2+bx+cax2+bx+c, with a=1a=1, b=-1b=1 and c=-1c=1.

This has solutions given by the quadratic formula

x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-1)+-sqrt((-1)^2 - (4xx1xx(-1))))/(2*1)x=b±b24ac2a=(1)±(1)2(4×1×(1))21

=(1+-sqrt(1+4))/2 = (1+-sqrt(5))/2=1±1+42=1±52

Method 2 - Completing the square

0 = x^2-x-1 = (x-1/2)^2-(1/2)^2-10=x2x1=(x12)2(12)21

= (x-1/2)^2-5/4=(x12)254

Add 5/454 to both ends to get:

(x-1/2)^2 = 5/4(x12)2=54

So:

x-1/2 = +-sqrt(5/4) = +-sqrt(5)/sqrt(4) = +-sqrt(5)/2x12=±54=±54=±52

Add 1/212 to both sides to get:

x = 1/2+-sqrt(5)/2 = (1+-sqrt(5))/2x=12±52=1±52

How are the two methods related?

0 = ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))0=ax2+bx+c=a(x+b2a)2+(cb24a)

Hence:

(x+b/(2a))^2 = (b^2/(4a)-c)/a = (b^2 - 4ac)/(4a^2)(x+b2a)2=b24aca=b24ac4a2

So:

x+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)x+b2a=±b24ac4a2=±b24ac4a2

=+-sqrt(b^2 - 4ac)/(2a)=±b24ac2a

Subtract b/(2a)b2a from both sides to get:

x = (-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a