Question

What are all the real zeros of y = (x - 12)^3 - 10?
A. x=3 square-root 10+12
B. x=3 square-root -10+12

Answer 1

i did this in a quick check last week to.... x.x

Answer 2

Do you know that answer? Because I got it wrong twice lol that's why there is only two options @Icedragon50

Answer 3

Can you use the equation editor to write the answers, I need to seem them clearer to give you the answer choice that is correct.

Answer 4

x=\[\sqrt[3]{10+12}\]
or
x=\[\sqrt[3]{-10+12}\]

Answer 5

@twistnflip

Answer 6

\[y=(x-12)^{3}-10\] is the equation, the other post are the answers @twistnflip

Answer 7

replace y with 0, and solve for x.

Answer 8

(when you are asked for "zeros" it is asking for x-interects....)

Answer 9

intercepts*

Answer 10

so its -10+12? @SolomonZelman

Answer 11

\(\large\color{black}{ \displaystyle y = (x - 12)^3 - 10 }\)

the very first step (set y=0)
\(\large\color{black}{ \displaystyle 0 = (x - 12)^3 - 10 }\)

Now, solving for x.
\(\large\color{black}{ \displaystyle 10 = (x - 12)^3 }\)
(I added 10 to both sides)

Answer 12

\(\large\color{black}{ \displaystyle \sqrt[3]{10} = x - 12 }\)

Answer 13

\(\large\color{black}{ \displaystyle \sqrt[3]{10}+12 = x }\)

Answer 14

Thank you very much! @SolomonZelman , I gave you a medal

Answer 15

I don't need medals, my profile looks suspicious because of their multiplicity\(\bf ....\)

The cube root, like any odd root is not going to have a ± sign, and this why we ended up with only 1 zero.