Question

Please Help!!!
(question below)

Answer 1

@tkhunny

Answer 2

\(x+3 = 0 \longrightarrow x = -3\)

It better be zero at x = -3. If it's zero at x = 3, something is very wrong.

This eliminates two. Now what?

Answer 3

so it's either C or D, right

Answer 4

?

Answer 5

You were not paying attention. Those are the ones that were ruled out.

Repeat: "If it's zero at x = 3, something is very wrong. "

Where are C and D zero? It look like x = 3 to me.

Answer 6

oh sorry

Answer 7

so A or B, I get it

Answer 8

so how do I know which one? @tkhunny

Answer 9

You look at it and think about it?

x = 0 should lead to the cube root of 3 (1.44ish). Do they do that.

Do you know any perfect cubes?

\(\sqrt[3]{1} = 1\)
\(\sqrt[3]{8} = 2\)
\(\sqrt[3]{27} = 3\)

Do they do that?

Answer 10

no, neither of them

Answer 11

Really? Try x = 5.

Answer 12

x = 5, y = 1.709...

Answer 13

wait, that's wrong

Answer 14

I don't get how to do that

Answer 15

I'm confused

Answer 16

@tkhunny

Answer 17

?? x = 5

\(\sqrt[3]{5+3} = \sqrt[3]{8} = 2\)

Perhaps you were using the SQUARE root?

Answer 18

oh yeah I was

Answer 19

so for x = 5, y = 125?

Answer 20

No, it's not \((5+3)^{3} = 512\). It's \(\sqrt[3]{5+3} = \sqrt[3]{8} = 2\)