Question Number 20115 by Tinkutara last updated on 22/Aug/17
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$
Answered by ajfour last updated on 22/Aug/17
$${roots}\:{are}\:{real}\:{and}\:{product}\:{of}\:{roots}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} −\mathrm{1}\:>\:\mathrm{0}\:\: \\ $$$${sum}\:{of}\:{roots}=−\frac{\mathrm{2}{a}}{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:>\mathrm{0} \\ $$$${knowing}\:{that}\:\mathrm{1}−{a}^{\mathrm{2}} \:<\:\mathrm{0}\: \\ $$$${we}\:{conclude}\:\:{a}>\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0}\:,{D}\geqslant\mathrm{0}\:,\:{f}\left(\mathrm{1}\right)<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{4}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\:\geqslant\:\mathrm{0}\:\:;\:{true}\:{for}\:{all}\:{a} \\ $$$$\:{f}\left(\mathrm{0}\right)=−\mathrm{1}\:<\mathrm{0}\:;\:{always}\:{true} \\ $$$$\:{f}\left(\mathrm{1}\right)=\mathrm{1}−{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}={a}\left(\mathrm{2}−{a}\right)\:<\:\mathrm{0} \\ $$$$\Rightarrow\:{a}<\mathrm{0}\:\:{or}\:{a}>\mathrm{2} \\ $$$${knowing}\:{a}^{\mathrm{2}} >\mathrm{1}\:{and}\:{a}>\mathrm{0} \\ $$$${we}\:{can}\:{say}\:\:{a}\in\left(\mathrm{2},\:\infty\right)\:. \\ $$
Commented by ajfour last updated on 22/Aug/17
$${yes},\:{thanks}. \\ $$
Commented by Tinkutara last updated on 22/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$