How do you solve using the completing the square method #2x^2-8x+3=0#?

1 Answer
May 6, 2016

Answer:

#x = 2+-sqrt(10)/2#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(2x-4)# and #b=sqrt(10)# as follows.

Multiply by #2# first to make the leading term a perfect square:

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

#=4x^2-16x+6#

#=(2x)^2-2(2x)(4)+6#

#=(2x-4)^2-16+6#

#=(2x-4)^2-10#

#=(2x-4)^2-(sqrt(10))^2#

#=((2x-4)-sqrt(10))((2x-4)+sqrt(10))#

#=(2x-4-sqrt(10))(2x-4+sqrt(10))#

#=(2(x-2-sqrt(10)/2))(2(x-2+sqrt(10)/2))#

#=4(x-2-sqrt(10)/2)(x-2+sqrt(10)/2)#

Hence:

#x = 2+-sqrt(10)/2#