Question

Radicals

Answer 1

\[\frac{ \sqrt[3]{5x} }{ \sqrt[3]{2x^2} }\]

Answer 2

How do you think you would approach this?

Answer 3

You can do something to get rid of that annoying radical

Answer 4

Not sure :thonk:

Answer 5

\[\frac{ \sqrt[3]{5x} }{ \sqrt[3]{2x^2} }\]
\[\frac{ \sqrt[3]{5} \times x^{\frac{ 1 }{ 3 }} }{ \sqrt[3]{2} \times x^{\frac{ 2 }{ 3 }}}\]
\[\frac{ \sqrt[3]{5} \times x ^{\frac{ 1 }{ 3 } - \frac{ 2 }{ 3 }}}{ \sqrt[3]{2}}\]
\[\frac{ \sqrt[3]{5} \times x^{-\frac{ 1 }{ 3 }} }{ \sqrt[3]{2} }\]
\[\frac{ \sqrt[3]{5} }{ \sqrt[3]{2} \times x^{1/3}}\]

Answer 6

I’m lost.. (sorry, dozed off..)

Answer 7

I’m confused by what those steps mean.

Answer 8

Which ones

Answer 9

I understand 1 and 2 but not after that.

Answer 10

Quotient of Powers:

\[\frac{ a^{m} }{ a^{n}} = a^{m - n}\]

Answer 11

Negative Exponent:

\[a^{-m} = \frac{ 1 }{ a^{m}}\]

Answer 12

Got I hate math... *sigh*

Answer 13

I don’t understand any of this... kms

Answer 14

\[\frac{ \sqrt[3]{5} \times x ^{\frac{ 1 }{ 3 } - \frac{ 2 }{ 3 }}}{ \sqrt[3]{2}}\]

You don't understand how I got here?

Answer 15

No 😔 but math has never been easy for me 😔

Answer 16

I just used the quotient of powers, and then since I got a negative exponent, I moved it down to the denominator.

Answer 17

Okay..

Answer 18

Completely ignored the 3rd root of 5 and 2 this whole time. Just focused on the x since there is nothing you can do about the coefficients.

Answer 19

Kinda makes sense..

Answer 20

Yeah, just convert them to exponents, separate them from the radicals, do the quotient of powers (division of exponents), subtract, get a negative exponent, do the negative exponent rule by moving it down to the denominator, and you're done.

Answer 21

I love the migraine that comes with math..

Answer 22

Think I got it.. thankfully there is only one of those.

Answer 23

Okay