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McStewart
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#1
Report Thread starter 12 years ago
#1
Just wondering if anyone can tell me the answers to this and how to get to them, me and a mate have been trying this for a while and got nowhere


x^2/3 - x^1/3 - 2 =0

thanks in advance
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AlphaNumeric
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Let y = x^{\frac{1}{3}}

Then you've a quadratic in y, solve that then work out x from that.
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vector
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I think I'm right in saying

x^(2/3) - x^(1/3) -2 = 0

x^(1/3) = cubic root of x

Let cubic root of x = ax

2ax - ax - 2 = 0
ax - 2 = 0

Therefore the cubic root of x = 2

Cube both sides

x = 2^3
x = 8
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AlphaNumeric
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#4
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(Original post by vector771)
Let cubic root of x = ax
Why?
(Original post by vector771)
2ax - ax - 2 = 0
That's not what you get, you should get a^{2}x^{2}-ax^{2}-2=0.

My method gives the quadratic y^{2}-y-2=0 so y = x^{\frac{1}{3}} = -1 \; , \; 2

So x = -1 or 2. Your method gets one of the right answers by doing two incorrect things.
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harr
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(Original post by vector771)
I think I'm right in saying

x^(2/3) - x^(1/3) -2 = 0

x^(1/3) = cubic root of x

Let cubic root of x = ax

2ax - ax - 2 = 0
ax - 2 = 0

Therefore the cubic root of x = 2

Cube both sides

x = 2^3
x = 8
I think you've made a mistake. I get
(ax)^{2} - ax - 2 = 0 not 2ax - ax - 2 = 0
Factorising this gives
(ax+1)(ax-2)=0
ax=2,-1
x=8,-1

Edit: beaten to it.
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UpliftMof0
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[QUOTE=AlphaNumeric]Why?
QUOTE]
I think he is just doing a similar thing to what you did AlphaNumeric, but using different notation (ax instead of y)

mebbe.!
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AlphaNumeric
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#7
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Okay, that explains the crap notation he uses (use a different symbol, not something involving x!) but not the doubling instead of squaring.
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*LEGEND*
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#8
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I dont think thats the end of it
So youve got y = 2 or -1
but y = cubed root of x
so x = 8 or -1
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harr
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#9
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Okay, that explains the crap notation he uses (use a different symbol, not something involving x!)
I agree that the notation isn't very user friendly, but it is no worse than several other notation styles I can think of (such as the f(x) notation). I'd definitely stick to using a different letter though if I was trying to work it out from scratch.
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