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

1 Answer
Oct 8, 2015

Answer:

The solutions are #4 \pm sqrt(13)#.

Explanation:

We must manipulate the expression into something of the form

#x^2 + 2ax+ a^2#

Since the term #2ax# equals #-8x#, it means that #a=-4#, and thus #a^2=16#. We need #13# more units for the constant to reach that value, so we can add and subtract:

#x^2 -8x+ 3 + 13 - 13 = 0 -> x^2 -8x +16 =13#

But now we have that #x^2 -8x +16 = (x-4)^2#, so the equation becomes

#(x-4)^2 = 13#

Taking square roots of both sides:

#x-4 = \pm \sqrt(13)#

#x=4 \pm \sqrt(13)#