solving with powers

12345678

Junior Member
Joined
Mar 30, 2013
Messages
102
hello, I've been having trouble with numerous questions similar to this one, so I was wondering if somebody could explain it step by step.
'Solve he equation: X^2/3 = 2X^1/3'
Now understand X^2/3 = Cube root( X)^2 and 2X^1/3 = cube root(2x)
My next line of thinking was to get rid of the cube roots, by cubing each side, so:
X^2 = 2X
Solving this to 0, I get X(X+2)=0
So X= 0 and X=-2
Substituting these values back into the equation, 0^2/3 = 2(0)^1/3 so 0 = 0, meaning 0 is a solution
However, I don't know I that answer was purely conidial as cube root (4) =Cube root (-4), when the answers are X = 0 and X = 8.
If anybody can explain where I have gone wrong, it would be greatly appreciated.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,876
hello, I've been having trouble with numerous questions similar to this one, so I was wondering if somebody could explain it step by step.
'Solve he equation: X^2/3 = 2X^1/3'
Now understand X^2/3 = Cube root( X)^2 and 2X^1/3 = cube root(2x)
My next line of thinking was to get rid of the cube roots, by cubing each side, so:
X^2 = 2X
Good idea- but you didn't cube each side! \(\displaystyle (x^{2/3})^3= x^3\) alright and \(\displaystyle (x^{1/3})^3= x\) but you forgot \(\displaystyle 2^3= 8\).

Cubing each side gives \(\displaystyle x^2= 8x\)

Solving this to 0, I get X(X+2)=0
And there is an error here. "Solving to 0", by subtracting 8x from both sides gives \(\displaystyle x^2- 8x= x(x- 8)= 0\), not \(\displaystyle x^2+ 8x= 0\)

So X= 0 and X=-2
Substituting these values back into the equation, 0^2/3 = 2(0)^1/3 so 0 = 0, meaning 0 is a solution
However, I don't know I that answer was purely conidial
I have no idea what "conidial" means.

as cube root (4) =Cube root (-4)
No, cube root of 4 is not equal to cube root of -4.

when the answers are X = 0 and X = 8.
If anybody can explain where I have gone wrong, it would be greatly appreciated.
Two errors as pointed out above.
 

12345678

Junior Member
Joined
Mar 30, 2013
Messages
102
Good idea- but you didn't cube each side! \(\displaystyle (x^{2/3})^3= x^3\) alright and \(\displaystyle (x^{1/3})^3= x\) but you forgot \(\displaystyle 2^3= 8\).

Cubing each side gives \(\displaystyle x^2= 8x\)


And there is an error here. "Solving to 0", by subtracting 8x from both sides gives \(\displaystyle x^2- 8x= x(x- 8)= 0\), not \(\displaystyle x^2+ 8x= 0\)


I have no idea what "conidial" means.


No, cube root of 4 is not equal to cube root of -4.


Two errors as pointed out above.
Simple error by my part. Thanks for the reply- and I didn't mean to spell conidial, it was a miss-spelling. Cheers again!
 

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,399
Simple error by my part. Thanks for the reply- and I didn't mean to spell conidial, it was a > > miss-spelling < < . Cheers again!
Just so you know, the correct spelling of "misspelling" does not contain a hyphen, and it has one fewer esses in it than what you typed. Cheers!
 
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