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Question Number 5936 by Master Tria last updated on 05/Jun/16
If the equation x^4 −4x^3 +ax^2 +bx+1=0 has four positive roots, then a, b are
$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+\mathrm{1}=\mathrm{0} \\ $$$$\:\mathrm{has}\:\mathrm{four}\:\mathrm{positive}\:\mathrm{roots},\:\mathrm{then}\:{a},\:{b}\:\mathrm{are} \\ $$
Answered by Yozzii last updated on 06/Jun/16
Consider the form of (x−t)(x−u)(x−r)(x−q)=0 for a general polynomial equation of degree 4, where t,u,r & q are its roots. (x^2 −(u+t)x+ut)(x^2 −(r+q)x+rq)=0 x^4 −(r+q+u+t)x^3 +(rq+ut+ur+uq+tq+tr)x^2 −(rqu+rqt+utr+utq)x+utrq=0 Let S=−(r+q+u+t),T=rq+ut+ur+uq+tq+tr J=−(rqu+rqt+utr+utq) and N=utrq. ⇒x^4 +Sx^3 +Tx^2 +Jx+N=0. The roots of (∗) x^4 −4x^3 +ax^2 +bx+1=0 are all positive. So for (∗), T>0 and J<0. But, T=a and J=b (S=−4, N=1). ⇒a>0 and b<0.
$${Consider}\:{the}\:{form}\:{of}\:\left({x}−{t}\right)\left({x}−{u}\right)\left({x}−{r}\right)\left({x}−{q}\right)=\mathrm{0} \\ $$$${for}\:{a}\:{general}\:{polynomial}\:{equation}\:{of}\:{degree}\:\mathrm{4}, \\ $$$${where}\:{t},{u},{r}\:\&\:{q}\:{are}\:{its}\:{roots}. \\ $$$$\left({x}^{\mathrm{2}} −\left({u}+{t}\right){x}+{ut}\right)\left({x}^{\mathrm{2}} −\left({r}+{q}\right){x}+{rq}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left({r}+{q}+{u}+{t}\right){x}^{\mathrm{3}} +\left({rq}+{ut}+{ur}+{uq}+{tq}+{tr}\right){x}^{\mathrm{2}} −\left({rqu}+{rqt}+{utr}+{utq}\right){x}+{utrq}=\mathrm{0} \\ $$$${Let}\:{S}=−\left({r}+{q}+{u}+{t}\right),{T}={rq}+{ut}+{ur}+{uq}+{tq}+{tr} \\ $$$${J}=−\left({rqu}+{rqt}+{utr}+{utq}\right)\:{and}\:{N}={utrq}. \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{Sx}^{\mathrm{3}} +{Tx}^{\mathrm{2}} +{Jx}+{N}=\mathrm{0}. \\ $$$$ \\ $$$${The}\:{roots}\:{of}\:\left(\ast\right)\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+\mathrm{1}=\mathrm{0} \\ $$$${are}\:{all}\:{positive}.\:{So}\:{for}\:\left(\ast\right),\:{T}>\mathrm{0}\:{and}\:{J}<\mathrm{0}. \\ $$$${But},\:{T}={a}\:{and}\:{J}={b}\:\:\left({S}=−\mathrm{4},\:{N}=\mathrm{1}\right). \\ $$$$\Rightarrow{a}>\mathrm{0}\:{and}\:{b}<\mathrm{0}. \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 09/Jun/16
x^4 −4x^3 +ax^2 +bx+1=0 −−−−−−−−−−−−−−−−− Number of +ve roots≤Number of variations of signs So a>0 and b<0 Otherwise number of variation of signs will be less than 4.
$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+\mathrm{1}=\mathrm{0} \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$${Number}\:{of}\:+{ve}\:{roots}\leqslant{Number}\:{of}\:{variations}\:{of}\:{signs} \\ $$$${So}\:\:\:{a}>\mathrm{0}\:\:\:{and}\:\:\:{b}<\mathrm{0} \\ $$$${Otherwise}\:{number}\:{of}\:{variation}\:{of}\:{signs}\:{will}\:{be} \\ $$$${less}\:{than}\:\:\mathrm{4}. \\ $$

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