Question

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

Answer 1

`(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`

Answer 2

`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`