anonymous one year ago 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

1. anonymous

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

2. anonymous

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

3. anonymous

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.

4. anonymous

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

5. anonymous

@twistnflip

6. anonymous

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

7. SolomonZelman

replace y with 0, and solve for x.

8. SolomonZelman

9. SolomonZelman

intercepts*

10. anonymous

so its -10+12? @SolomonZelman

11. SolomonZelman

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

12. SolomonZelman

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

13. SolomonZelman

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

14. anonymous

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

15. SolomonZelman

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.