Posted by **Jennifer** on Sunday, May 4, 2008 at 3:48pm.

Here's one of the questions I have problems with:

cube root of (x+2)=6th root of (9x+10)

A. -1, 6

B. [-13 +/- sqrt(193)]/2

C. 1, -6

D. [13 +/- sqrt(145)]/2

Please explain how to solve this. Should you cube both sides, or raise both to the 6th? and I get lost trying to do more than square a radical. i'd show my "work" which is basically nonsense but it's useless because I honestly have no idea how to solve this.

thanks in advance for the help. having an example of how to solve these kinds of problems would help me a lot, so if you can step by step would be great.

- math-radical equations -
**Jennifer**, Sunday, May 4, 2008 at 3:54pm
sorry to post twice; computer meltdown...

- math-radical equations -
**Damon**, Sunday, May 4, 2008 at 4:56pm
(x+2)^(1/3) = (9x+10)^(1/6)

raise both sides to the sixth

[ (x+2)^(1/3) ] ^6 = (x+2)^2

[ (9x+10)^(1/6)]^6 = (9x+10)^1 = 9x+10

so

x^2 + 4 x + 4 = 9 x + 10

x^2 -5 x - 6 = 0

(x-6)(x+1) = 0

x = 6 or x = -1

- math-radical equations -
**Damon**, Sunday, May 4, 2008 at 5:04pm
Oh, and always check the answers

in this case you have for x = -1

1^1/3 = 1^1/6 yes, ok

and for x = 6

8^1/3 = 64^1/6 ? = (8*8)^1/6 =(2*2*2*2*2*2)^1/6 = 2

2 = 2 sure enough

- math-radical equations -
**Damon**, Sunday, May 4, 2008 at 5:07pm
LOL, if I had really noticed it was multiple choice I would have tried x = -1 immediately :)

- math-radical equations -
**Jennifer**, Sunday, May 4, 2008 at 9:24pm
thanks, Damon - you are awesome =)

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