How do you solve #x^2-12x+11=0# by completing the square?

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Answer 1 · 2017-04-17T14:11:36

#x={1,11}#

#x^2-12x+11=0#
#x^2-12xcolor(red)(+25-25)+11=0#

#x^2-12x+36color(red)(-25)=0#
#x^2-12+36=(x-6)^2#
#25=5^2#

#(x-6)^2-5^2=0#

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

#(x-6)^2-5^2=(x-6-5)(x-6+5)=(x-11)(x-1)#

#(x-11)(x-1)=0#

#"Either (x-11) or (x-1) is equal to 0"#

#"if "(x-11)=0" then "x=11#
#"if "(x-1==0" then "x=1#