How do you solve #x^2 -16x+12=0# by completing the square?

1 Answer
Jim G.
Dec 17, 2017

#x=8+-sqrt52#

Explanation:

#"using the method of "color(blue)"completing the square"#

#• " ensure the coefficient of the "x^2" term is 1"#

#• " add "(1/2"coefficient of x-term")^2" to both sides"#

#x^2-16x+12=0larr" coefficient of "x^2" term is 1"#

#x^2+2(-8)xcolor(red)(+64)+12=0color(red)(+64)#

#rArr(x-8)^2+12=64#

#rArr(x-8)^2=52#

#color(blue)"take the square root of both sides"#

#sqrt((x-8))^2=+-sqrt52larrcolor(blue)"note plus or minus"#

#rArrx-8=+-sqrt52#

#"add 8 to both sides"#

#rArrx=8+-sqrt52larrcolor(red)"exact values"#

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