Question

Could I have some help with this Algebra 2 question? I am not for sure how to do this and every number I come up with makes no sense. 3 (^5sqrt x+2^3)+3 =27

Answer 1

can you show how you are getting your numbers?

Answer 2

I subtracted by three then I divided 24 by 3 and got 8

Answer 3

^5sqrt x+2^3, this is not very readable

perhaps you mean: 5rt(x+2^3) ? or is it: 5rt(x) + 2^3 ?

Answer 4

after that I got confused what happened next

Answer 5

2rt = sqrt
3rt = cbrt
4rt
5rt
etc ...

these are easier to read then trying the ^n sqrt

Answer 6

the first one you typed was right! I couldn't figure out how to type it.

Answer 7

3k+3 =27

3k =24

k = 8

then unroot it

k^(1/5) = 8^(1/5) assuming k is a 5rt radical

Answer 8

or if k = 5rt(x) + 2^3 then we continue by subtracting 8 and then worry about the rt

Answer 9

\[3\sqrt[5]{x+2^{3}}+3=27\]

Answer 10

I get stuck at 5rt(x+2)^3 = 8

Answer 11

which is it:
\[a)~~~\sqrt[5]{x+(2)^3}\]or
\[b)~~~\sqrt[5]{(x+2)^3}\]

Answer 12

a

Answer 13

\[\sqrt[5]{x+8}=8\]
\[(\sqrt[5]{x+8})^{1/5}=8^{1/5}\]
\[x+8=8^{1/5}\]
and continue

Answer 14

pfft, ^5 not ^1/5

Answer 15

for example:

\[\sqrt[n]{b}=b^{1/n}\]
if we ^n it we get
\[(b^{1/n})^n=b^{n/n}=b\]

Answer 16

so, x+8 = 8^5 and continue

Answer 17

ok, so it's ok that the answer ends up being a crazy number? 32760?

Answer 18

another way to prolly see it better is:

\[\sqrt[5]{k}=8~~\implies~~k=8^5\]

Answer 19

yes, big numbers are allowed to be solutions :)

Answer 20

x = 8^5 - 8 is a find solution

3(5rt(8^5-8 +8)) + 3 = 27

3*5rt(8^5) + 3 = 27

3*8 + 3 = 27 is true

Answer 21

thank you so much for not just giving me the answers and actually helping me understand how this works!

Answer 22

youre welcome