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3x^2/3= 6x

How to solve this?

What I did was divide by 3 to give me
X^2/3=2x
Now I'm stuck

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Original post by Ggffffhh
3x^2/3= 6x

How to solve this?

What I did was divide by 3 to give me
X^2/3=2x
Now I'm stuck

i mean i got x=0 but i'm sure that's not the answer you're looking for xD
i'm stuck too
:s-smilie:
:/
@Zacken
Reply 2
Avatar for Zacken
Zacken
22
Original post by Ggffffhh

X^2/3=2x
Now I'm stuck


Original post by thefatone
i mean i got x=0 but i'm sure that's not the answer you're looking for xD
i'm stuck too
:s-smilie:
:/
@Zacken


Simples: x2/3−2x=0⇒x(x−1/3−2)=0x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0 so either x=0x=0 or x−1/3=2x^{-1/3} = 2 the latter is easy to solve, since x−1/3=2=81/3=(18)−1/3x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}.
There is another solution as well as x=0x=0 and that is x=18x=\dfrac{1}{8}. This comes from dividing by x23x^{\frac{2}{3}}.
(edited 8 years ago)
Original post by Zacken
Simples: x2/3−2x=0⇒x(x−1/3−2)=0x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0 so either x=0x=0 or x−1/3=2x^{-1/3} = 2 the latter is easy to solve, since x−1/3=2=81/3=(18)−1/3x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}.


oh this step caught me

x(x−13−2)=0x\left(x^\frac{-1}{3} -2\right)=0

i just took out the whole x xD and everything i knew about indices went out the window
(edited 8 years ago)
Reply 5
Avatar for Zacken
Zacken
22
Original post by thefatone
oh this step caught me

x(x−13−2)=0x\left(x^-\dfrac{1}{3} -2\right)=0


I think you mean x^{-\frac{1}{3}}, use the curly brackets.
Reply 6
Avatar for Zacken
Zacken
22
Original post by Bath~Student
false..


Oh yeah? Elaborate.
Original post by Zacken
Oh yeah? Elaborate.


Not inherently incorrect.

But I feel your method is rather awkward.

Why not cube both sides, factor out an x^2 giving x^2 ( 8x -1 ) = 0 so x=0 or x=1/8..
Original post by Zacken
I think you mean x^{-\frac{1}{3}}, use the curly brackets.


oh ok it just looked a little weird when i edited it
x−13x^{-\frac{1}{3}}
Reply 9
Avatar for Zacken
Zacken
22
Original post by Bath~Student
Not inherently incorrect.

But I feel your method is rather awkward.

Why not cube both sides, factor out an x^2 giving x^2 ( 8x -1 ) = 0 so x=0 or x=1/8..


It's entirely correct.

Because I don't want to.
Original post by Bath~Student
Not inherently incorrect.

But I feel your method is rather awkward.

Why not cube both sides, factor out an x^2 giving x^2 ( 8x -1 ) = 0 so x=0 or x=1/8..


Well then saying 'false' is completely unjustified.
But the quickest method is noting that x=0x=0 is a solution, then dividing by x23x^{\frac{2}{3}} to get the other one immediately.
Original post by Zacken
It's entirely correct.

Because I don't want to.


It is.

But it is beyond ugly factoring out an x and having the reciprocal of x to the third power in your brackets.
Reply 12
Avatar for sabana
sabana
20
After that step x^2 = 6x
So x^2 - 6x=0
Now you have a quadratic
x(x -6)=0
x=0 or x =6
Original post by IrrationalRoot
Well then saying 'false' is completely unjustified.
But the quickest method is noting that x=0x=0 is a solution, then dividing by x23x^{\frac{2}{3}} to get the other one immediately.


I fully concur. Why one would want the reciprocal of x to the third power within the factoring brackets is beyond me. But then I am not a future snob-university mathematician, so what do I know?
Original post by sabana
After that step x^2 = 6x
So x^2 - 6x=0
Now you have a quadratic
x(x -6)=0
x=0 or x =6


errrrr
Reply 15
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Zacken
22
Original post by sabana
After that step x^2 = 6x
So x^2 - 6x=0
Now you have a quadratic
x(x -6)=0
x=0 or x =6


The OP means 3x^(2/3) = 6x. :smile:
Reply 16
Avatar for sabana
sabana
20
Original post by Zacken
The OP means 3x^(2/3) = 6x. :smile:


Ooops my mistake sorry
Reply 17
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Zacken
22
Original post by sabana
Ooops my mistake sorry


No worries, I can see why you'd be confused though, don't worry about it! :biggrin:
Original post by Zacken
No worries, I can see why you'd be confused though, don't worry about it! :biggrin:


What a philanthropist.
Original post by Zacken
It's entirely correct.

Because I don't want to.


Original post by Bath~Student
It is.

But it is beyond ugly factoring out an x and having the reciprocal of x to the third power in your brackets.


Have to agree with Bath Student, there is much simpler way to solve this not involving the reciprocal and third power, just confusing OP otherwise.

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