How do you solve #x^2- 1/4x - 3/64=64# by completing the square?

1 Answer
Oct 21, 2017

Answer:

#x=(1+-10sqrt(41))/8#.

Explanation:

To complete the square, you can apply the formula #x^2+ax=(x+1/2a)^2-(1/2a)^2#.

You have to substitute #a=(-1/4)/2=-1/8# to solve this problem.

#x^2-1/4x-3/64=64#
#x^2-1/4x=64+3/64#
#(x-1/8)^2-(-1/8)^2=64+3/64#
#(x-1/8)^2=64+3/64+(-1/8)^2#
#(x-1/8)^2=64+4/64=1025/16#

Notice that #1025=5^2*41#.
The root of the equation is
#x-1/8=+-sqrt(1025/16)#
#x-1/8=+-(5sqrt(41))/4#
#x=(1+-10sqrt(41))/8#.