give exact and approximate solutions to three decimal places: y^2-8y+16=1
Subtract 1 from both sides to get 0 on the right side:
You can factor that as
Set each equal to 0:
But since you mentioned decimal approximating that
made me think you were supposed to do it using
the quadratic formula:
You would use
for
,
for
,
for
and
for
x = (8±2)/2
Using the +,
Using the -,
Edwin