How do you complete the square to solve #x^2 -4x -5=0#?

Question

57554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-25T00:37:10

#0 = (x^2 - 4x) - 5 = (x^2 -4x + 4) -4 -5 = (x-2)^2-9 quad # so

#(x-2)^2 = 9 or x-2 = \pm \sqrt{9} or x = 2\pm 3# so

# x=-1 or x=5#

Answer 2 · 2018-04-25T00:45:36

#-1=x=5#

We can regroup the equation as:

#(x^2-4x)-5=0#

This equation is in the form:

#ax^2+bx+c=0# where #a=1#. (That is important.)

We divide #b# by two.

#b=-4#

#=>b/2=-2# square the answer.

#=>(-2)^2=4#

We add this answer inside the parenthesis

#=>(x^2-4x+4)-5=0# Since we add four to the original equation, we need to adjust it by subtracting it by four.

#=>(x^2-4x+4)-5-4=0#

#=>(x^2-4x+4)-9=0#

Now, note that #x^2-4x+4# is in the form

#a^2+2ab+b^2# where #a=x# and #b=-2#

We can rewrite this into the form #(a+b)^2#

#=>(x-2)^2-9=0#

#=>(x-2)^2=9#

#=>(x-2)=+-sqrt9# don't forget the #+-#!

#=>x=+-3+2#

#=>-3+2=x=3+2#

#=>-1=x=5#