I would appreciate if somebody could help me with the following problem
Q: Quadratic Equation $x^4+ax^3+bx^2+ax+1=0$ have four real roots $x=\frac{1}{\alpha^3},\frac{1}{\alpha},\alpha,\alpha^3(\alpha>0)$ and $2a+b=14$.
Find $a,b=?(a,b\in\mathbb{R})$
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1Why don't you just expand out the product of the linear factors? – meta_warrior Dec 14 '13 at 09:13
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1This is quartic and not quadratic. You could modify the title. – Claude Leibovici Dec 14 '13 at 09:14
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Hint:
$$0=x^4+ax^3+bx^2+ax+1=(x-\alpha)(x-\alpha^3)\left(x-\frac1\alpha\right)\left(x-\frac1{\alpha^3}\right)$$
Now compare coefficients in both sides (Viete's Formulas), for example:
$$-a=\alpha+\alpha^3+\frac1\alpha+\frac1{\alpha^3}\;,\;\;b=\alpha^4+\frac1{\alpha^4}+\alpha^2+\frac1{\alpha^2}+2\;\ldots$$
DonAntonio
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have you found any elegant solution of the bi-quadratic equation in $\alpha+1/\alpha$? or any other way to find $a,b$ without solving the equation? – lab bhattacharjee Dec 14 '13 at 09:41
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I've not even tried it. Hopefully the above will help, but it could be it won't. – DonAntonio Dec 14 '13 at 09:42
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Solving for the condition 2a+b=14 leads to a quite unpleasant equation for the OP. The resulting equation reduces to the product of a linear term squared (one root), multiplied by a quadratic term (two distinct roots) , multiplied by a quartic term; this last one has no real root. Any suggestion ? – Claude Leibovici Dec 14 '13 at 09:57
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Shall we let the OP work his way into his problem, with these and those hints? Maybe he'll succeed, or else he'll write back. – DonAntonio Dec 14 '13 at 09:59
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@DonAntonio. I totally agree with you. What I was concern in my comment is this extra condition which makes the problem quite difficult (I suppose) except if there is a smart transform in terms of some combination of alpha. Cheers. – Claude Leibovici Dec 14 '13 at 10:22
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lab has a point. if we write $x=\alpha +\frac1{\alpha}$ then we get $$ -ax + b = x^2+2 $$ or $$ x= \frac12 \left((-a \pm \sqrt{a^2-4(2-b)}\right) $$ but for a 'nice' problem the $2a+b=14$ should simplify the surd, and it doesn't quite seem to. maybe my arithmetic is wrong, or maybe i'm just a hopeless optimist with a penchant for erroneous oversimplification. (well i know that, actually)
David Holden
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