# Proof polynomial has only one real root.

I need to prove that this polynomial equation: $$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2}).$$ has only one root. That it has one real root is obvious because it is of odd degree. But Descartes rules here fails to bound the number of roots to one.

• For $a=0$ it has a triple root at $x=0,\,$ i.e. there are three real roots. – gammatester Apr 4 '14 at 9:00
• @gammatester True, but I think OP is not counting multiplicity of roots (that is, there is still only one root, $0$). – 5xum Apr 4 '14 at 9:03
• Sorry, my mistake...I forgot a condition on $a$ that now has been added. – Ambesh Apr 4 '14 at 9:04
• en.wikipedia.org/wiki/Sturm%27s_theorem – athos Dec 16 '14 at 2:34

Consider $$p(x) = x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a$$ First, we note that if $x< 0$, each term is negative, hence there are no negative roots. Also, $p(0) = -a < 0$. Further,

$$p'(x) = 5x^4-4(3-a)x^3+3(3-2a)x^2-2ax+2a$$

So it is sufficient to show that $p'(x) > 0$ for $x > 0, \; a \in (0, \frac12)$.

For this, note that by AM-GM inequality, $\frac12 ax^2+2a \ge 2ax$, so it is sufficient to show that: $5x^4+\frac12(18-13a)x^2 > 4(3-a)x^3$. By AM-GM we again have: $$5x^4+\tfrac12(18-13a)x^2 \ge 2\sqrt{\frac{5(18-13a)}2}x^3$$

So it is enough to show $5(18-13a) > 8(3-a)^2 \iff 8a^2+17a < 18$, which is true for $a \in (0, \frac12)$.

• The first time you apply AM-MG, shouldn't one of the $a$'s be rooted or squared? From that point onwards I'm lost. I don't see from where comes the next expression. – Ambesh Apr 4 '14 at 10:49
• @Ambesh You're correct, I missed that. Edited that. – Macavity Apr 4 '14 at 10:55

I try to factorize your formula :

$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2})$

$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=x^5+(3-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a$

$=x^5+(1-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a+2x^{3}-2x^{4}$
$=x^5-1+1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2ax+2ax^{2}-a+2x^{3}-2x^{4}$ $=x^5-1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2a(x-1)+2x^{3}(1-x)+1+a+2ax^{2}$ $=(x-1)(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a+2ax^{2}$$=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a-(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2}-2ax^{2})$ $=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})-(x^{4}+x^{2}+x+ax^{3}-ax^{2}+a+2x^{2}-2ax^{2})$

$\Longrightarrow{x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)=a}$

$\Longrightarrow{f(x)=x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)}$

$\Longrightarrow{f^{'}(x)=0}$

$\Longrightarrow{-10x^{4}=2-4a>0}$

a contradiction shows that $x=0$ is its unique root!