Question

what are all the real zeroes y=(x-12)^3-7

Answer 1

help @Hero

Answer 2

please!

Answer 3

Hello???????

Answer 4

Hint: set y = 0 and solve for x

Answer 5

so 0=(x-12)^3-7
0=x^3-12^3-7
0=x^3-1728-7
0=x^3-1721

Answer 6

is this right

Answer 7

@Hero is this going on the right track

Answer 8

HELLLOOOO???? ANYONE!!! I NEED HELP!

Answer 9

@yummydum can u help

Answer 10

Hello? @yummydum

Answer 11

i don't think hero do the expansion of (x-12)^3 right.....should move the number to the other side and take cube root of both side

Answer 12

yummydum, why do you use the method of finding roots for quaratic equation to solve a third degree equation?

Answer 13

can u help @tamtoan

Answer 14

i mean this is a pretty icky equation they've got here

Answer 15

@lala2, i just typed earlier , as hero said, set y = 0 then you have (x-12)^3 - 7 = 0.

==> (x-12)^3 = 7
(x-12) =

then continue from here

Answer 16

whats after that

Answer 17

@yummydum , you should never use the formula of solving quadratic equation and apply it to the third degree equation unless in some special case you can reduce to second degoree equation, otherwise, the answer will never be right

Answer 18

im confused

Answer 19

and you can't solve this ?