Solve for \(r:\) \(\frac{1}{r}\)+\(\frac{1}{r^{2}}\)+\(\frac{1}{r^{3}}\)+\(\frac{1}{r^{4}}\)\(=340\)

I was only able to solve r=14r=\frac{1}{4} but WolframAlpha has r0.221r ≈ -0.221, which I don't know how to get.

Note by Ryan Merino
1 month, 1 week ago

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Cardano's method

Pi Han Goh - 1 month, 1 week ago

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Neat! I haven't worked much with cubics, obviously. I didn't understand the end of the second example though:

y=st=10+63310+633=2y = s-t = \sqrt[3]{10 + 6\sqrt{3}} - \sqrt[3]{-10 + 6\sqrt{3}} = 2

Is there something I'm missing?

David Stiff - 1 month, 1 week ago

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8t3t3=20 \dfrac8{t^3} - t^3 = 20

(2tt)3+3(2tt)2tt=20 (\tfrac2t - t)^3 + 3(\tfrac2t - t)\cdot \tfrac2t \cdot t = 20

(2tt)3+6(2tt)=20 (\tfrac2t - t)^3 + 6(\tfrac2t - t) = 20

Let z=2ttz = \tfrac2t - t , then z3+6z=20 z^3 + 6z = 20

Rational root theorem tells us that z=2z=2 is a root and the other two roots are not real.

so z=2tt=2z = \tfrac2t - t = 2 only.

Or st=2s - t = 2 only.

Pi Han Goh - 1 month, 1 week ago

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@Pi Han Goh Interesting...the wiki didn't include any of those steps. Would you mind detailing how you went from the first equation to the second? Thanks!

David Stiff - 1 month, 1 week ago

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@David Stiff Use a3b3=(ab)3+3ab(ab) a^3 - b^3 = (a - b)^3 + 3ab(a - b)

Pi Han Goh - 1 month, 1 week ago

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@Pi Han Goh Cool. Thanks!

David Stiff - 1 month, 1 week ago

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Thanks. Didn't know this method 'til now.

Ryan Merino - 1 month ago

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Hmm... seems complicated. :)

David Stiff - 1 month, 1 week ago

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