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minusequals

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- Thread starter minusequals
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minusequals

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- #2

Number Nine

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By letting u = x^{2} you get a cubic in u. While it is possible to solve cubics such as this by hand, you likely have better things to do. You could solve it numerically or use a program such as Maple to write out an explicit formula. Here's the exact answer given by Maple for the positive value of x where they intersect (hard to read and parse):

(1/6)*sqrt(6)*sqrt((972+12*sqrt(6573))^(1/3)*((972+12*sqrt(6573))^(2/3)-12))/(972+12*sqrt(6573))^(1/3)

or approximately 1.385703836.

(1/6)*sqrt(6)*sqrt((972+12*sqrt(6573))^(1/3)*((972+12*sqrt(6573))^(2/3)-12))/(972+12*sqrt(6573))^(1/3)

or approximately 1.385703836.

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- #4

HallsofIvy

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Well, I don't! Using the cubic formula (and, I will admit, a calculator) I got 1.9195 as a solution to the cubic, to four decimal places. The square root of that is 1.3855, basically your answer. Using that you can reduce the problem to a quadratic.By letting u = x^{2}you get a cubic in u. While it is possible to solve cubics such as this by hand, you likely have better things to do.

LCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.You could solve it numerically or use a program such as Maple to write out an explicit formula. Here's the exact answer given by Maple for the positive value of x where they intersect (hard to read and parse):

(1/6)*sqrt(6)*sqrt((972+12*sqrt(6573))^(1/3)*((972+12*sqrt(6573))^(2/3)-12))/(972+12*sqrt(6573))^(1/3)

or approximately 1.385703836.

- #5

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While it is possible to solve cubics such as this by hand, you likely have better things to do...

Well, I don't!

Heh. Nothing else to do eh?

LCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.

"With great power comes great responsibility." -- Spiderman

- #6

Number Nine

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Heh. Nothing else to do eh?

"With great power comes great responsibility." -- Spiderman

Pfft...he was only quoting his Aunt. Your knowledge of Marvel comics is insufficient for you to be giving advice on this forum.

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