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How many roots? (Posted on 2011-07-31) Difficulty: 3 of 5
Prove that the equation
has at most 2 real number solutions.

No Solution Yet Submitted by Ady TZIDON    
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Solution Solution (spoiler) - I think | Comment 1 of 6
The roots of a polynomial always come in complex conjugate pairs.  Therefore there are zero, two or four real roots.  Assume the general case that the roots are a+-bi and c+-di.  The product of all the roots must equal -1.  Therefore -1 = a^2 + b^2 +c^2 +d^2.  All real numbers have a square>=0, therefore at least one of the terms a,b,c or d must be imaginary, and therefore at least 2 must be imaginary.  Therefore there can be a maximum of only 2 real roots.

  Posted by Kenny M on 2011-08-02 17:48:55
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