How do you solve #x^2 - 8x + 3 = 0#?
1 Answer
Apr 30, 2016
Explanation:
Complete the square and use the difference of squares identity:
#a^2-b^2=(a-b)(a+b)#
with
#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)#
