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Prove that the equation $x^{15} + 7x^{3} - 5 = 0$ has exactly one real solution.

A hint that has been given by the teacher is to analyze the function $f(x) = x^{15} + 7x^{3} - 5$.

Rodrigo de Azevedo
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Martiiin
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3 Answers3

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Hints:

Do what your teacher says, and observe:

(1) The equation is a polynomial one of odd degree and thus it has at least one real solution;

(2) $\;f'(x)=15x^{14}+21x^2\;$ is a non-negative function, thus...

DonAntonio
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    Why couldn't you just use the rule of signs? – Parcly Taxel Oct 22 '16 at 12:24
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    First, because I think it is way easier and simpler this way (though not more elementary...). Second, the rule of the signs is pretty unwidely unknown: even many university students haven't heard of it. Third, in this case this rule only serves to deduce the existence of a positive root, yet no info for negative ones as the number of sign changes of $;f(-x);$ is zero... – DonAntonio Oct 22 '16 at 12:30
  • Since there are no sign changes in $f(-x)$, we can conclude there are no negative real roots at all. (What were you expecting, $-2$ negative roots?) I've written up an answer that uses the rule. – Parcly Taxel Oct 22 '16 at 12:41
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If $0\le a<b$ show that $a^{15}<b^{15}$, and similarly $7a^3<7b^3$.

So $a^{15}+7a^3 < b^{15}+7b^3$. So the function $g(x)=x^{15}+7x^3$ is an increasing function. Check that $g(0) = 0$ and $g(1)=8$. So $g$'s values change from $0$ to $8$ when $x$ varies from 0 to 1. Being continuous and increasing there is exactly an $x_0\in(0,1)$ such that $g(x_0)=5$. That $x_o$ is the root for $f(x)$. As $g$ is increasing it cannot take the value 5 again. SO that is the only root.

P Vanchinathan
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We will use Descartes's rule of signs to determine the number of positive and negative roots of $x^{15}+7x^3-5$. (Zero is obviously not a root.)

  1. The non-zero coefficients of the polynomial per se change sign exactly once. Therefore, it has exactly one positive root.
  2. Replacing $x$ with $-x$ we get $-x^{15}-7x^3-5$, which has no sign changes at all. Therefore the original polynomial has no negative roots.

In conclusion, the original polynomial has exactly one real root, and it is positive.

Parcly Taxel
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