Question Number 17359 by ajfour last updated on 04/Jul/17
$$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m},\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}+\mathrm{m}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\mathrm{3m}\right)\mathrm{x}+\left(\mathrm{1}+\mathrm{8m}\right)=\mathrm{0}\:; \\ $$$$\:\:\:\mathrm{m}\:\in\:\mathrm{R}\:,\:\mathrm{has}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{positive}\:? \\ $$
Commented by ajfour last updated on 04/Jul/17
$$\mathrm{book}'\mathrm{s}\:\mathrm{answer}: \\ $$$$\:\mathrm{m}\in\:\left(−\infty,\:−\mathrm{1}\right)\:\cup\:\left[\mathrm{3},\infty\right) \\ $$$$\mathrm{my}\:\mathrm{answer}\:\mathrm{differed}.. \\ $$
Commented by ajfour last updated on 04/Jul/17
$$\mathrm{thanks}\:\mathrm{Sir}.\:\mathrm{very}\:\mathrm{precise}. \\ $$
Commented by prakash jain last updated on 04/Jul/17
![roots are ((−b)/(2a))±((√(b^2 −4ac))/(2a)) −(b/(2a)) lies in between roots condition for +ve roots>0 −(b/(2a))≥0,αβ>0,D≥0 (((1+3m))/((1+m)))≥0⇒m∈(−∞,−1)∪[−(1/3),∞) (((1+8m))/((1+m)))>0⇒m∈(−∞,−1)∪(−(1/8),∞) D=4{(1+3m)^2 −(1+m)(1+8m)}≥0 ⇒m(m−3)≥0⇒m∈(−∞,0]∪[3,∞) answer m∈(−∞,−1)∪(−(1/8),0]∪[3,∞)](https://www.tinkutara.com/question/Q17365.png)
$$\mathrm{roots}\:\mathrm{are}\:\frac{−\mathrm{b}}{\mathrm{2a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$−\frac{{b}}{\mathrm{2}{a}}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{between}\:\mathrm{roots} \\ $$$$\mathrm{condition}\:\mathrm{for}\:+\mathrm{ve}\:\mathrm{roots}>\mathrm{0} \\ $$$$−\frac{\mathrm{b}}{\mathrm{2a}}\geqslant\mathrm{0},\alpha\beta>\mathrm{0},{D}\geqslant\mathrm{0} \\ $$$$\frac{\left(\mathrm{1}+\mathrm{3}{m}\right)}{\left(\mathrm{1}+{m}\right)}\geqslant\mathrm{0}\Rightarrow{m}\in\left(−\infty,−\mathrm{1}\right)\cup\left[−\frac{\mathrm{1}}{\mathrm{3}},\infty\right) \\ $$$$\frac{\left(\mathrm{1}+\mathrm{8}{m}\right)}{\left(\mathrm{1}+{m}\right)}>\mathrm{0}\Rightarrow{m}\in\left(−\infty,−\mathrm{1}\right)\cup\left(−\frac{\mathrm{1}}{\mathrm{8}},\infty\right) \\ $$$${D}=\mathrm{4}\left\{\left(\mathrm{1}+\mathrm{3}{m}\right)^{\mathrm{2}} −\left(\mathrm{1}+{m}\right)\left(\mathrm{1}+\mathrm{8}{m}\right)\right\}\geqslant\mathrm{0} \\ $$$$\Rightarrow{m}\left({m}−\mathrm{3}\right)\geqslant\mathrm{0}\Rightarrow{m}\in\left(−\infty,\mathrm{0}\right]\cup\left[\mathrm{3},\infty\right) \\ $$$${answer} \\ $$$${m}\in\left(−\infty,−\mathrm{1}\right)\cup\left(−\frac{\mathrm{1}}{\mathrm{8}},\mathrm{0}\right]\cup\left[\mathrm{3},\infty\right) \\ $$
Commented by Tinkutara last updated on 05/Jul/17
$$\mathrm{ajfour}\:\mathrm{Sir},\:\mathrm{which}\:\mathrm{book}\:\mathrm{is}\:\mathrm{that}?\:\mathrm{Can}\:\mathrm{you} \\ $$$$\mathrm{please}\:\mathrm{tell}\:\mathrm{its}\:\mathrm{description}\:\mathrm{so}\:\mathrm{that}\:\mathrm{I}\:\mathrm{can} \\ $$$$\mathrm{prepare}\:\mathrm{well}\:\mathrm{from}\:\mathrm{it}? \\ $$
Commented by ajfour last updated on 05/Jul/17
$$\mathrm{S}.\mathrm{K}.\:\mathrm{Goyal}'\mathrm{s}\:\mathrm{Algebra},\:\mathrm{Arihant} \\ $$$$\mathrm{Prakashan}. \\ $$