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Finding parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$

has two distinct positive roots

Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$

So $\displaystyle \bigg(x+\frac{1}{x}\bigg)^2+a\bigg(x+\frac{1}{x}\bigg)-8=0$

put $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t\;,$ where $|t|\geq 2$

so $t^2+at-8=0.$ So for real roots $D\geq 0.$ So $a^2+32\geq 0$

So $\displaystyle t = \frac{-a\pm \sqrt{a^2+32}}{2}.$

could some help me how to solve it, thanks

2 Answers 2

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Your strategy is good. The equation $t^2+at-8=0$ has two real roots, namely $$ r_1=\frac{-a+\sqrt{a^2+32}}{2} \qquad r_2=\frac{-a-\sqrt{a^2+32}}{2} $$ Note that $r_1>0$ and $r_2<0$. Consider now $$ x+\frac{1}{x}=r_i $$ which becomes $$ x^2-r_ix+1=0 $$ If we take $i=2$, we see that the equation either has no real root or it has two negative real roots. So we need to take $i=1$. The equation has (distinct) positive roots if and only if its discriminant is positive: $$ r_1^2-4>0 $$

This becomes$$(\sqrt{a^2+32}-a)^2-16>0$$and, doing some simplifications,$$a^2+8>a\sqrt{a^2+32}$$This is true whenever $a<0$. If $a\ge0$ we can square and get$$a^4+16a^2+64>a^4+32a^2$$that is, $a^2<4$, which gives $0\le a<2$. Thus the final solution is $a<2$.

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    $\frac{1}{4} \, \left(a^{2} - 2a\sqrt{a^{2} + 32} + (a^{2} + 32)\right) - 4 > 0$ – 2017-02-27
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    is equivalent to – 2017-02-27
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    $a^{2} - a\sqrt{a^{2} + 32} > -8$ – 2017-02-27
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    $a^{2} + 8 > a\sqrt{a^{2} + 32}$ – 2017-02-27
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    $a^{2} < 4$ ... – 2017-02-27
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    $-2 < a < 2$. ... – 2017-02-27
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    @AgalnamedDesire I left it to the OP. Your computation has a mistake. – 2017-02-27
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    I was careless with the case $a < 0$. – 2017-02-27
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    @AgalnamedDesire Yes, a common mistake. `;-)` Always be careful when squaring! – 2017-02-27
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    Here is a more interesting question regarding the same quartic equation. For what value of $a$ does $x^{4} +ax^{3} - 6x^{2} + ax + 1 = 0$ have exactly one positive solution? Your solution provided here adapts to this. It does not use any Calculus - you can use the Cauchy-Schwartz Inequality to assert that $x + 1/x \geq 2$. – 2017-02-28
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    @AgalnamedDesire That's easy: $a=2$ – 2017-02-28
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With $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t$ the roots occur when $t^2+at-8=0$.

$f(x)=x^4+ax^3-6x^2+ax+1 = 0$ shows $f(0)=1>0$ and $\displaystyle \lim_{x\to+\infty} f=+\infty$, which show $f$ has two distinct positive roots if there is $x_0>0$ for which $f(x_0)<0$. Let $\displaystyle \bigg(x_0+\frac{1}{x_0}\bigg)=t_0\geq2$ and $f(x_0)=t_0^2+at_0-8<0$ or $$4+2a-8\leq t_0^2+at_0-8<0$$ Shows $a<2$.