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  #1  
Old 11-25-2002, 06:13 AM
Jim Brier Jim Brier is offline
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Default Math Conundrum

If the quantity [x + (2/x)]is squared and set equal to 6, what is the quantity [(x cubed) + ((8)/(x cubed))]?
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  #2  
Old 11-25-2002, 09:47 AM
irchans irchans is offline
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Default A guess

zero
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  #3  
Old 11-25-2002, 10:44 AM
marbles marbles is offline
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Default Re: Math Conundrum

Taking the square root of both sides and multiplying through by x, you can solve for x by the quadratic.
X=square root of 6 plus or minus square root of 14 all over 2. Plugging both results into your new equation, the result is approximately 29.39388 either way.
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Old 11-25-2002, 01:25 PM
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Default Re: Math Conundrum

180
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  #5  
Old 11-25-2002, 01:30 PM
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Default Re: Math Conundrum

(x + 2/x)^2 = 6
(x + 2/x) = + - sqrt(6)
x^2 + 2 = + - (sqrt(6))*x
x^2 + - (sqrt(6))x + 2 = 0

x = ( + - sqrt(6) + - sqrt(6 MINUS 4*1*2) ) / 2

x = (+ - sqrt(6) + - sqrt(-2)) / 2

x is a complex number, however that doesn't mean the problem doesn't have an answer.
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Old 11-25-2002, 01:38 PM
marbles marbles is offline
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Default Re: Math Conundrum

The first step of the quadratic is -b, not +-b. It results in two possible answers for X, but they both give the same answer when plugged into the next equation.
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Old 11-25-2002, 01:41 PM
marbles marbles is offline
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Default Scratch that...

Oh, I see your point... I was assuming that we are only allowing the + square root of 6. Allowing for negative square root of 6 does make the problem a little more messy.
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  #8  
Old 11-25-2002, 01:46 PM
pudley4 pudley4 is offline
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Default Re: Math Conundrum

No, if you use the quadratic equation to solve for X, you get an imaginary number, not a real number.

(x+2/x)^2=6
x + 2/x = sqrt(6)
x^2 - sqrt(6)x + 2 = 0

plug into the quadratic equation [-b +/- sqrt(b^2 - 4ac)]/2a

notice b^2 - 4ac = 6 - 8 = -2, so you end up with an imaginary number.

[edit] and I see this point has already been made [img]/forums/images/icons/smile.gif[/img] [/edit]
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  #9  
Old 11-25-2002, 01:53 PM
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Default Re: Math Conundrum

Sorry, I missed the "is squared" in my first reply.
Imagine that!
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  #10  
Old 11-25-2002, 01:54 PM
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Default solution

I am using the notations a^b = a to the b power, and sqrt(c) = the positive square root of c.

Expanding out (x + (2/x))^3, we get x^3 + 6x + 12/x + 8/(x^3).

Rearranging the terms,

(x + (2/x))^3 = x^3 + 8/(x^3) + 6(x + 2/x)

Now, we are given (x + 2/x)^2 = 6. Therefore (x + 2/x) = sqrt(6) or (x + 2/x) = - sqrt(6).

so, we have (sqrt(6))^3 = x^3 + 8/(x^3) + 6*sqrt(6), or
................(-sqrt(6))^3 = x^3 + 8/(x^3) - 6*sqrt(6)

solving each: 6*sqrt(6) = x^3 + 8/(x^3) + 6*sqrt(6)
0 = x^3 + 8/(x^3)

-6*sqrt(6) = x^3 + 8/(x^3) - 6*sqrt(6)
0 = x^3 + 8/(x^3)

In either case, x^3 + 8/(x^3) = 0, which is what we were asked to find. Looks like irchans is correct, it all disappears into nothingness.
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