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Reply 20

sabana

Original post by Zacken

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

Lol yeah, that actually makes the qn a lot more difficult. Don't think I recall doing a question like that before.

Reply 21

Ggffffhh

Original post by Zacken

Simples:

x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0

so either

x=0

x^{-1/3} = 2

the latter is easy to solve, since

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

Yeaaaah fully get it now!
Thankyou so much

Reply 22

Zacken

Original post by Ggffffhh

Yeaaaah fully get it now!
Thankyou so much

No worries, glad that helped. :smile:

Reply 23

Zacken

Original post by Ggffffhh

Yeaaaah fully get it now!
Thankyou so much

Original post by TheTopStudent

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.

Yep, defo confused the OP. I don't really see the difference - they are logically equivalent, it's just applying the inverse at a later time - mine allows you to observe the solution from a straightforward injectivity argument which is why I prefer it.

But hey, why don't you post your 'alternative' method for others to benefit from! :smile:

(edited 7 years ago)

Reply 24

Bath~Student

Original post by Zacken

Cos I already did. Those smiley faces are painfully obnoxious..

Reply 25

Zacken

Original post by Bath~Student

Cos I already did. Those smiley faces are painfully obnoxious..

Reply 26

hellomynameisr

Original post by Zacken

Simples:

x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0

so either

x=0

x^{-1/3} = 2

the latter is easy to solve, since

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

Could you explain

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

and how x^-1/3 = 2 = 8^1/3?

Reply 27

Zacken

Original post by hellomynameisr

Could you explain

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

and how x^-1/3 = 2 = 8^1/3?

2 is the same thing as

8^{1/3}

. So if I say that

x^{-1/3} = 2

then it must mean that

x = 8^{-1}

Reply 28

Bath~Student

Original post by hellomynameisr

Could you explain

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

and how x^-1/3 = 2 = 8^1/3?

It isn't. It is, however, a stylistic abomination.

Reply 29

hellomynameisr

Original post by Zacken

2 is the same thing as

8^{1/3}

. So if I say that

x^{-1/3} = 2

then it must mean that

x = 8^{-1}

Ohhh I see. Thank you :smile:

Reply 30

Zacken

Original post by hellomynameisr

Ohhh I see. Thank you :smile:

No worries. :smile:

Reply 31

celiacrowther

I multiplied both sides by 3 which gave me 3x^2=18x, then divided both sides by x which gave me 3x=18 and then solved for x which gave me x=6

Reply 32

Zacken

Original post by celiacrowther

I multiplied both sides by 3 which gave me 3x^2=18x, then divided both sides by x which gave me 3x=18 and then solved for x which gave me x=6

The question is 3x^(2/3) not (3x^2)/3. :smile:

Reply 33

Notnek

Original post by Zacken

Simples:

x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0

so either

x=0

x^{-1/3} = 2

the latter is easy to solve, since

x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

It's obviously personal preference but I like your way the best.

A method without dividing the equation or cubing both sides feels nicer to me. Factorising is underrated.