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December 20, 2014

December 20, 2014

Posted by **ayisha** on Thursday, October 13, 2011 at 9:13am.

- math -
**Steve**, Thursday, October 13, 2011 at 10:22amx^4 + 4x^2 + m + 8/x^2 + 4/x^4

well, we will have

(x^2 + k + 2/x^2)^2

x^4 + kx^2 + 2 + + kx^2 + k^2 + 2k/x^2 + 2 + 2k/x^2 + 4/x^4

x^4 + 2kx^2 + 4 + k^2 + 4k/x^2 + 4/x^4

equate like powers of x to find that we must have k=2

so m = 4+k^2 = 8

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