How do you solve #x^2 - 4x + 7 =0# by completing the square?

1 Answer
Aug 10, 2017

#x = 2+-sqrt(3)i#

Explanation:

Given:

#x^2-4x+7=0#

While completing the square we will find that this takes the form of the sum of a square and a positive number. As a result it has no solution in real numbers, but we can solve it using complex numbers.

The imaginary unit #i# satisfies #i^2=-1#

The difference of squares identity can be written:

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

We can use this with #a=(x-2)# and #b=sqrt(3)i# as follows:

#0 = x^2-4x+7#

#color(white)(0) = x^2-4x+4+3#

#color(white)(0) = (x-2)^2+(sqrt(3))^2#

#color(white)(0) = (x-2)^2-(sqrt(3)i)^2#

#color(white)(0) = ((x-2)-sqrt(3)i)((x-2)+sqrt(3)i)#

#color(white)(0) = (x-2-sqrt(3)i)(x-2+sqrt(3)i)#

Hence the two roots are:

#x = 2+sqrt(3)i" "# and #" "x = 2-sqrt(3)i#