Question

Can someone help me solve a radical equation?

Answer 1

I'll type the equation out

Answer 2

\[2\sqrt[3]{10-3x}=\sqrt[3]{2-x}\]

Answer 3

to undo a cube root, you "cube" it (raise it to the third power, or "use an exponent of 3"

Answer 4

but wouldn't it be 1/3 if I raise it?

Answer 5

so "cube" both sides:
\[ (2\sqrt[3]{10-3x})^3=(\sqrt[3]{2-x})^3\]

if we use ⅓ power instead of the radical sign, it would be like this
\[ ( 2 (10-3x)^\frac{1}{3})^3 = ((2-x)^\frac{1}{3})^3 \]

Answer 6

does x=9?

Answer 7

if you know about this "rule"
\[ \left(a^b\right)^c = a^{b\cdot c}\]
we can simplify the expression:
\[ ( 2 (10-3x)^\frac{1}{3})^3 = ((2-x)^\frac{1}{3})^3 \\
( 2^3 (10-3x)^\frac{3}{3}) = ((2-x)^\frac{3}{3})\]

Answer 8

oh wait, i did't cube root the 2

Answer 9

i didn't distribute correctly

Answer 10

notice the 3/3 is 1, and anything to the 1 power is itself.

yes, you have to cube "both sides" which on the left includes the 2

Answer 11

so that would give me \[2^3*10-3x=2-x\]

Answer 12

so now you have
\[ 2^3 (10-3x) = 2-x \]
2^3 is 8
so
\[ 8(10-3x)_ = 2-x \\
80 -24x = 2-x\]

Answer 13

oh okay so I would have to disturb 8 to 10-3x

Answer 14

yes.
I am not getting a very pretty number for x.
78 = 23 x
x= 78/23

Answer 15

after working it out some, will it be 78=23x ?

Answer 16

yes, and then 78/23 = x
ugly!

Answer 17

and then you divide them alright

Answer 18

It can't be simplified

Answer 19

no, that is as simple as it gets. you can change to a decimal and round it but that is not exact.

Answer 20

And how would I check my work with this answer?

Answer 21

\[2\sqrt[3]{10-3(\frac{ 78 }{ 23})}=\sqrt[3]{2-(\frac{ 78 }{ 23})}\]

Answer 22

yes, a calculator shows it works

Answer 23

i am not allowed to use a calculator

Answer 24

My teacher doesn't allow it

Answer 25

no, but I can...

Answer 26

Alright, thanks anyways. I'll figure out how to check it. :)