Question

4(3x-3)^2/3=36

Answer 1

I can start by taking 4 from each side so it becomes (3x-3)^2/3=9 but from here I'm lost

Answer 2

I know the answers should be 10 and -8 but I don't know how to get there

Answer 3

I have to use Rational exponents to find the answer and x should = 10 and -8

Answer 4

@ganeshie8

Answer 5

Please note that if I multiply both sides of your original equation by 3, I get:
\[4{\left( {3x - 2} \right)^2} = 108\]
next I divide both sides, by 4, and I can write:

\[{\left( {3x - 2} \right)^2} = 27\]

Answer 6

finally I take the square root of both sides:
\[3x - 2 = \pm 3\sqrt 3 \]

Answer 7

:o...i guess i used the lengthy method :p

Answer 8

nevertheless it is correct @SyedMohammed98

Answer 9

ok that just made me more lost i guess since they are showing us a different way. hum

Answer 10

ill wait and ask the teacher since it her way of teaching that is confusing me

Answer 11

please note that you have to solve these two equations:
\[\begin{gathered}
3x - 2 = + 3\sqrt 3 \hfill \\
3x - 2 = - 3\sqrt 3 \hfill \\
\end{gathered} \]

Answer 12

I think i have that i just need to do it they way they are showing since i have to show my work i think i almost have it

Answer 13

I'm sorry, since I have made an error please I rewrite those equation which you have to solve:
\[\begin{gathered}
3x - 3 = + 3\sqrt 3 \hfill \\
3x - 3 = - 3\sqrt 3 \hfill \\
\end{gathered} \]

Answer 14

Is the question \(4(3x-3)^{2/3} = 36\)?

Answer 15

yes

Answer 16

I have made an error @Jhannybean

Answer 17

\[(3x-3)^{2/3} = 9 \implies \sqrt[3]{(3x-3)^2} =9\]\[\left(\sqrt[3]{(3x-3)^2}\right)^3 =(9)^3\]\[(3x-2)^2 = 9^3\]\[3x-2=9^{3/2}\]etc....

Answer 18

ok thank you that looks right

Answer 19

\[9^{3/2} = 27\]

Answer 20

\[x=\frac{27+2}{3}\]

Answer 21

please note that we have:
\[3x - 3 = \pm 27\]

Answer 22

Oh yeah, whoops.

Answer 23

In my earlier step, i forgot the \(\pm\).. \[3x-2 = \pm 9^{3/2}\]

Answer 24

I got it now thank you

Answer 25

Np