Question Number 198932 by necx122 last updated on 25/Oct/23
$${Find}\:{the}\:{value}\:{of}\:{m}\:{given}\:{that}\:{the} \\ $$$${roots}\:{of}\:{x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{3}} +\mathrm{70}{x}^{\mathrm{2}} −\mathrm{120}{x}+{m}=\mathrm{0} \\ $$$${form}\:{a}\:{geometric}\:{progression}. \\ $$
Answered by AST last updated on 25/Oct/23
$${m}={abcd}=\left({br}^{−\mathrm{1}} \right){b}\left({br}\right)\left({br}^{\mathrm{2}} \right)={b}^{\mathrm{4}} {r}^{\mathrm{2}} \\ $$$${a}+{b}+{c}+{d}=\mathrm{15}\Rightarrow{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} \right)=\mathrm{15}…\left({i}\right) \\ $$$${ab}+{bc}+{cd}+{da}+{ca}+{bd}=\mathrm{70} \\ $$$$\Rightarrow{a}^{\mathrm{2}} \left({r}+{r}^{\mathrm{3}} +{r}^{\mathrm{5}} +{r}^{\mathrm{3}} +{r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right)=\mathrm{70}…\left({ii}\right) \\ $$$${abc}+{bcd}+{acd}+{abd}={b}^{\mathrm{3}} \left(\mathrm{1}+{r}^{\mathrm{3}} +{r}^{\mathrm{2}} +{r}\right)=\mathrm{120}…\left({iii}\right) \\ $$$$\left({i}\right)\&\left({iii}\right)\Rightarrow\frac{\mathrm{120}}{{b}^{\mathrm{3}} }=\frac{\mathrm{15}}{{a}}=\mathrm{8}{a}={b}^{\mathrm{3}} \\ $$$${b}^{\mathrm{3}} ={a}^{\mathrm{3}} {r}^{\mathrm{3}} =\mathrm{8}{a}\Rightarrow{a}^{\mathrm{2}} {r}^{\mathrm{3}} =\mathrm{8} \\ $$$$\Rightarrow{m}={b}^{\mathrm{4}} {r}^{\mathrm{2}} ={a}^{\mathrm{4}} {r}^{\mathrm{6}} =\left({a}^{\mathrm{2}} {r}^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{64} \\ $$