Question Number 18432 by Joel577 last updated on 21/Jul/17
$$\mathrm{Find}\:\mathrm{interval}\:{p}\:\mathrm{so} \\ $$$$\left({p}\:−\:\mathrm{2}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{p}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{negative}\:\mathrm{roots} \\ $$
Answered by 433 last updated on 21/Jul/17
$$ \\ $$$$ \\ $$$$\Delta\geqslant\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}\left({p}−\mathrm{1}\right)\left({p}−\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{4}\left({p}^{\mathrm{2}} −{p}^{\mathrm{2}} +\mathrm{3}{p}−\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{3}{p}−\mathrm{2}\geqslant\mathrm{0}\:\Rightarrow\:{p}\geqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\left(\mathrm{1}\right) \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} >\mathrm{0}\:\&\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} <\mathrm{0} \\ $$$$\frac{{p}−\mathrm{1}}{{p}−\mathrm{2}}>\mathrm{0}\:\&\:−\frac{\mathrm{2}{p}}{{p}−\mathrm{2}}<\mathrm{0} \\ $$$$\left({p}−\mathrm{1}\right)\left({p}−\mathrm{2}\right)>\mathrm{0}\:\&\:\mathrm{2}{p}\left({p}−\mathrm{2}\right)>\mathrm{0} \\ $$$${p}\in\left(−\infty,\mathrm{1}\right)\cup\left(\mathrm{2},+\infty\right)\:\&\:{p}\in\left(−\infty,\mathrm{0}\right)\cup\left(\mathrm{2},+\infty\right)\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right),\left(\mathrm{2}\right) \\ $$$${p}\in\left(\mathrm{2},+\infty\right) \\ $$
Commented by Joel577 last updated on 25/Jul/17
$${thank}\:{you}\:{very}\:{much} \\ $$