How do you solve #y^2-12y=-35# by completing the square?

2 Answers
May 9, 2018

Answer:

#(y-6)^2-1=0#

Explanation:

#y^2-12y+35-0#

#(y-6)^2+a=0#

#y^2-12y+36+a=y^2-12y+35#

#a=-1#

#(y-6)^2-1=0#

May 9, 2018

Answer:

#y=5" or "y=7#

Explanation:

#"to solve using "color(blue)"completing the square"#

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

#rArry^2+2(-6)ycolor(red)(+36)=-35color(red)(+36)#

#rArr(y-6)^2=1#

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

#sqrt((y-6)^2)=+-sqrt1larrcolor(blue)"note plus or minus"#

#rArry-6=+-1#

#"add 6 to both sides"#

#rArry=6+-1#

#rArry=6-1=5" or "y=6+1=7#