Question 631402
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If *[tex \LARGE a_0x^n\ +\ a_1x^{n-1}\ +\ a_2x^{n-2}\ +\ \cdots\ +\ a_{n-1}x\ +\ a_n\ \equiv\ 0] then *[tex \LARGE a_0\ =\ a_1\ =\ a_2\ =\ \cdots\ =\ a_{n-1}\ =\ a_n\ =\ 0].


So set *[tex \LARGE k^2\ -\ 3k\ +\ 2\ =\ 0], *[tex \LARGE k^2\ -\ 5k\ +\ 4\ =\ 0], and *[tex \LARGE k^2\ -\ 6k\ +\ 5\ =\ 0], and solve each.  The three quadratics will have a common zero and this is your value of *[tex \LARGE k]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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