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How many real roots does the equation e^(6x) - e^(4x) - 2e^(3x) - 12e^(2x) + e^x + 1 = 0 have?

1. 2
2. 4
3. 6
4. 1

Correct answer: 2

Solution

Setting t = e^x > 0, the polynomial f(t) = t⁶ - t⁴ - 2t³ - 12t² + t + 1 has exactly two sign changes for t > 0 (one between t=0 and t=1, and one between t=2 and t=3), giving exactly 2 positive real values of t and therefore 2 real values of x.

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