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Find number of roots of the equation $e^x(x^4 + 4x^3 + 12x^2 + 24x + 24) + 1 = 0$

Using Descartes rule, number of positive roots is zero and there can be a maximum of 4 negative roots.

Also, for the function $P(x)=x^4 + 4x^3 + 12x^2 + 24x + 24$, the double derivative $P''(x)>0$. So the function can have either 2 negative roots or no root at all.

But I still lack information required to prove that the function has no real roots.

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    $\begingroup$ But Descartes' rule applies to polynomials, and this is no polynomial. How do you know that you're allowed to use it? $\endgroup$ – Arthur Nov 17 '15 at 12:53
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    $\begingroup$ I am using it only for P(x). $\endgroup$ – Aditya Dev Nov 17 '15 at 13:01
  • $\begingroup$ I think it's enough to show that the polynomial is a positive function because e^x is already greater than zero. $\endgroup$ – Aditya Dev Nov 17 '15 at 13:02
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Just observe that

\begin{align} P(x) &=(x^4+4x^3+4x^2)+(8x^2+24x+18)+6\\ &=(x^2+2x)^2+2(2x+3)^2+6>0 \end{align}

Therefore $$e^xP(x)+1>0$$ for all $x \in \mathbb R$.

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  • $\begingroup$ This works because of the choice of coefficients. What if they were different? $\endgroup$ – Aditya Dev Nov 17 '15 at 13:04
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    $\begingroup$ @AdityaDev Technically, any positive univariate polynomial can be expressed as such a sum of squares. That's not the only way either, you may for e.g. say $P(x) > (x+1)^4+6(x+\frac16)^2$. Of course, unless the coefficients are "nice" you are unlikely to encounter it as a class room / contest problem. $\endgroup$ – Macavity Nov 17 '15 at 14:36
  • $\begingroup$ Is there any general way for expressing a polynomial like that? $\endgroup$ – Aditya Dev Nov 17 '15 at 14:44

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