Question

help please

Answer 1

@johnweldon1993

Answer 2

Okay ...our first step...is to divide both sides by 3...because we want to get that cubed root by itself on the left side...so we can do the same trick we did in the last problem...

So dividing both sides by 3 we have

\[\large \sqrt[3]{x^2} = 9\]

Right?

Answer 3

yes

Answer 4

\[\large \frac{ = 3\sqrt[3]{x^2}=27}{3} => \sqrt[3]{x^2} = 9\]

So now...we do the trick we did last time

Answer 5

whats that?

Answer 6

It was just me writing out as an equation of what dividing by 3 looks like :) exactly what I said above (just done visually)

Answer 7

so now what

Answer 8

So now...we have the cubed root alone on the left side of the equation...just like that last question...So in order to cancel out a cubed root...we raise it to the 3rd power.......just like to get rid of a square root we would raise to the 2nd power...and a 4th root we would raise to the 4th etc...

Answer 9

so confused haha..

Answer 10

So to cancel that out....we raise this all to the 3rd power....this will cancel out that cued root....

\[\large (\sqrt[3]{x^2})^3 = 9^3\]

This becomes

\[\large x^2 = 9^3\]

And 9^3 is just 9x9x9 = 729

So

\[\large x^2 = 729\]

Now what?

Answer 11

would u divied it by 2 hahha just a guess

Answer 12

Well let me clarify first...I think what you're getting confused on is the roots and powers thing...so hang on

Answer 13

\[\huge (\sqrt[\cancel{2}]{x^2})^\cancel{2} = x^2\]
\[\huge (\sqrt[\cancel{3}]{x^2})^\cancel{3} = x^2\]
\[\huge (\sqrt[\cancel{4}]{x^2})^\cancel{4} = x^2\]

Make sense? when you raise a root...to its power you cancel it...raising a cubed (2) root to the 2nd power...cancels it

Answer 14

ok yah i think it does now

Answer 15

Okay so above again...all we did was raise the cubed root to the third power so it canceled...

\[\huge (\sqrt[\cancel{3}]{x^2})^\cancel{3} = 9^3\]

Which is now

\[\huge x^2 = 729\]

Okay ready for the last step?

Answer 16

x = 27?

Answer 17

Beat me to it :) Yeah that's your answer

Answer 18

hahah thank you :)

Answer 19

Anytime!