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The set S has elements of the form 2n, with n taking on all natural number values. Let m be any element of S. If p is the number of real roots of the equation xm + 1 - x = 0, how many different values are possible for p?
    Correct answer is '1'. Can you explain this answer?
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    The set S has elements of the form 2n, with n taking on all natural nu...
    Let m - 2n for some natural number n.
    Consider the equation,
    xm+1- x = 0
     x(xm - 1) = 0
    x = 0 or xm = 1
     m is even,
    the possible real roots of the equation are, -1, 0 and 1.
    Since p = 3 for all n and m, there is only one possible value of p. Answer: 1
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    Most Upvoted Answer
    The set S has elements of the form 2n, with n taking on all natural nu...
    Understanding the Set S
    - The set S consists of elements of the form 2n, where n is a natural number.
    - This means that S = {2, 4, 6, 8, ...}.

    Finding the Number of Real Roots
    - Let m be any element of S, so m = 2k for some natural number k.
    - Substituting m into the equation xm + 1 - x = 0 gives (2k)x + 1 - x = 0.
    - Simplifying this equation, we get 2kx - x + 1 = 0, which further simplifies to (2k - 1)x + 1 = 0.

    Determining the Number of Real Roots
    - The equation (2k - 1)x + 1 = 0 is a linear equation.
    - A linear equation has only one root in the real number system.
    - Therefore, the equation xm + 1 - x = 0 has only one real root for any element m in set S.

    Conclusion
    - The number of possible values for p, the number of real roots of the equation, is 1.
    - This is because no matter which element m of set S is chosen, the equation will always have only one real root.
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    The set S has elements of the form 2n, with n taking on all natural number values. Let m be any element of S. If p is the number of real roots of the equation xm+ 1 - x = 0, how many different values are possible for p?Correct answer is '1'. Can you explain this answer?
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