MHB No Real Zeroes of $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$

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Prove that the polynomial $x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}$ has no real zeroes.
 
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Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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