How do you solve #x^2 - 8x + 3 = 0#?

1 Answer
Apr 30, 2016

Answer:

#x = 4+-sqrt(13)#

Explanation:

Complete the square and use the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(x-4)# and #b=sqrt(13)# as follows:

#0 = x^2-8x+3#

#=(x-4)^2-16+3#

#=(x-4)^2-13#

#=(x-4)^2-(sqrt(13))^2#

#=((x-4)-sqrt(13))((x-4)+sqrt(13))#

#=(x-4-sqrt(13))(x-4+sqrt(13))#

So:

#x = 4+-sqrt(13)#