Question
give exact and approximate solutions to three decimal places
y^2-12y+36=4
y^2-12y+36=4
Answers
y^2 - 12y + 32 = 0
(y-8)(y-4) = 0
the "exact" solutions are
x = 8 or x=4
(y-8)(y-4) = 0
the "exact" solutions are
x = 8 or x=4