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- Thread starter minusequals
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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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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.

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

Heh. Nothing else to do eh?Well, I don't!

"With great power comes great responsibility." -- SpidermanLCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.

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Pfft...he was only quoting his Aunt. Your knowledge of Marvel comics is insufficient for you to be giving advice on this forum.Heh. Nothing else to do eh?

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