How do you solve x^2 - x - 1 = 0x2−x−1=0?
1 Answer
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-1x2−x−1 is of the formax^2+bx+cax2+bx+c , witha=1a=1 ,b=-1b=−1 andc=-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±√b2−4ac2a=−(−1)±√(−1)2−(4×1×(−1))2⋅1
=(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=x2−x−1=(x−12)2−(12)2−1
= (x-1/2)^2-5/4=(x−12)2−54
Add
(x-1/2)^2 = 5/4(x−12)2=54
So:
x-1/2 = +-sqrt(5/4) = +-sqrt(5)/sqrt(4) = +-sqrt(5)/2x−12=±√54=±√5√4=±√52
Add
x = 1/2+-sqrt(5)/2 = (1+-sqrt(5))/2x=12±√52=1±√52
How are the two methods related?
Hence:
(x+b/(2a))^2 = (b^2/(4a)-c)/a = (b^2 - 4ac)/(4a^2)(x+b2a)2=b24a−ca=b2−4ac4a2
So:
x+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)x+b2a=±√b2−4ac4a2=±√b2−4ac√4a2
=+-sqrt(b^2 - 4ac)/(2a)=±√b2−4ac2a
Subtract
x = (-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a