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If the equation $x^4-4x^3+ax^2+bx+1=0$ has four positive roots then $a=\,?$ and $b=\,?$

$\textbf{A.}\,6,-4$

$\textbf{B.}\,-6,4$

$\textbf{C.}\,6,4$

$\textbf{D.}\,-6,-4$

we can replace options and check answers .. are there any other shortcuts we can use

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    $\begingroup$ Try the binomial theorem. $\endgroup$ – copper.hat Dec 20 '13 at 18:33
  • $\begingroup$ @copper.hat But what if we don't know there is only one solution? $\endgroup$ – Jean-Claude Arbaut Dec 20 '13 at 18:48
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    $\begingroup$ @arbautjc: Well, it is a 4th order monic polynomial. It is completely determined by its roots. $\endgroup$ – copper.hat Dec 20 '13 at 18:55
  • $\begingroup$ @copper.hat Understood! :-) $\endgroup$ – Jean-Claude Arbaut Dec 20 '13 at 18:57
  • $\begingroup$ The problem is badly formulated. I would prefer it if the condition "and $|a|=6, |b|=4$" were added. $\endgroup$ – Hagen von Eitzen Dec 20 '13 at 19:27
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$$(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$ $$=x^4-x^3\left(\sum_{4\ge i\ge1} x_i\right)+x^2\left(\sum_{4\ge i>j\ge1}x_ix_j\right)-x\left(\sum_{4\ge i>j>k\ge1}x_ix_jx_k\right)+x_1x_2x_3x_4 $$

If $x_i>0,1\le i\le4$ the coefficient of $x^2$ must be $>0$ and that of $x$ must be $<0$

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$$x^4-4x^3+6x^2-4x+1=(x-1)^4$$ $$a=6,b=-4$$

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By the Descartes rule of signs, you need the maximal number of four sign variations. The converse is not true, even with four sign variations, there may be one or two pairs of complex roots. But of the four possibilities given, the last three are ruled out because they give only two sign variations, allowing for only two or zero positive real roots.

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