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How can I prove that an equation has exactly 2 real solutions using some kind of real-analysis methods?

For example I have to prove it for $x^{16}+7x^2-5=0$.

f(x)=x^16+7x^2-5

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There is a zero of derivative between any two solutions. As $(x^{16} + 7x^2 - 5)^\prime = 16x^{15} + 14x = x\cdot(16x^{14} + 14)$, the only zero of derivative on real line is at $x = 0$, so the equation has at most two solutions (would there be at least three derivative would have at least two zeros).

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Descartes' rule of sign can be used to determine the number of real zeros.

"It tells us that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The number of negative real zeroes of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number."-MathPlanet

Link to website

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Hint: Take the derivative $f'$

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