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.
6 answers
sorry to post twice; computer meltdown...
(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
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
it really helps to write 15th root of x for example as x^(1/15) power
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
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
LOL, if I had really noticed it was multiple choice I would have tried x = -1 immediately :)
thanks, Damon - you are awesome =)