Question

Correction!!! HELP!!!!!!!!!! Need to find the exact and approximate solution in three decimal places.
PLease show me your work too. I am LOST!
y²-16y+64=25

Answer 1

factor
(y-8)^2=25
take the square root of both sides
y-8=5
or
y-8=-5
y=13 or y=3

Answer 2

@Kira, how did you do that thing? the screenshot? it's so cool

Answer 3

What would be the 3 decimal answer

Answer 4

@Alex I'm using a MacBook and the application equation editor. So I type my solution in it and pressed [Command], [Shift], and [3] all at one go before I cropped it...

Answer 5

Put in form of quadratic solution \[ax^2 + bx + c = 0\]
\[y^2-16y+64-25 = 0\]
\[a=1, b= -16, c= 39\]

Now use solution to find values:

Answer 6

\[y = (-b - \sqrt(b^2-4ac))/2a\]\[y = (-b + \sqrt(b^2-4ac))/2a\]

Plugging in the values, we get \[y = (-(-16) + \sqrt((-16)*(-16)-4*1*39)/2 = (16 + \sqrt(256-156))/2 = 13\]
\[y=(-(-16) - \sqrt((-16)*(-16)-4*1*39)/2 = (16 - \sqrt(256-156))/2 = 3\]

Factoring works if you can see the factoring, but this always works, even in real life where the problems don't all have integer coefficients :-)

Answer 7

@Kira, you can take a snapshot of just a section of your screen by doing Command-Shift-4 and sweeping out the desired rectangle

Answer 8

:)

Answer 9

Thank you!

Answer 10

My (or should I say our) pleasure to be of service.... :)