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#1
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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
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A guess
zero
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#3
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A proof to follow...
He's right, it is zero [img]/forums/images/icons/smile.gif[/img]
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#4
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Re: A proof to follow...
The easiest way is to factor the equation:
x^3 + 8x^(-3) becomes [x+2x^(-1)] * [x^2 - 2 + 4x^(-2)] If you notice, the second factor can be found from the initial equation: (x + 2/x)^2 = 6 [x^2 + 4 + 4x^(-2)] = 6 [x^2 - 2 + 4x^(-2)] = 0 Since this factor = 0, the equation equals 0 |
#5
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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. |
#6
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Re: Math Conundrum
180
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#7
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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. |
#8
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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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#9
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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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#10
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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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