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Question-124791

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Question Number 124791 by ajfour last updated on 06/Dec/20
Commented by ajfour last updated on 06/Dec/20
If p, q, r are roots of equation x^3 −x−c = 0 , find k.
$${If}\:{p},\:{q},\:{r}\:{are}\:{roots}\:{of}\:{equation} \\ $$$${x}^{\mathrm{3}} −{x}−{c}\:=\:\mathrm{0}\:\:\:,\:\:{find}\:{k}. \\ $$
Answered by ajfour last updated on 08/Dec/20
m=tan θ=−(r/k) ((2m)/(1−m^2 ))=tan 2θ=((p−r)/k) (−km)^3 =(−km)+c ....(I) (p/k)=((2m−m+m^3 )/(1−m^2 ))=((m(m^2 +1))/(1−m^2 )) p=((km(m^2 +1))/(1−m^2 )) k^3 m^3 (m^2 +1)^3 =(1−m^2 )^2 km(m^2 +1) +c(1−m^2 )^3 (km−c)(m^2 +1)^3 =(1−m^2 )^2 km(1+m^2 ) +c(1−m^2 )^3 4km^3 (m^2 +1)=c{(1−m^2 )^3 +(1+m^2 )^3 } ⇒ 4km^3 (1+m^2 )=2c(1+3m^4 ) ...(II) from (I)&(II) 4km^3 (1+m^2 )=2km(1−k^2 m^2 )(1+3m^4 ) ⇒ 2m^2 (1+m^2 )=(1−k^2 m^2 )(1+3m^4 ) ⇒ 2m^2 +2m^4 =1+3m^4 −k^2 m^2 (1+3m^4 ) ⇒ k^2 m^2 =((m^4 −2m^2 +1)/(1+3m^4 )) = (((1−m^2 )^2 )/(1+3m^4 )) & k^2 m^2 =((c^2 (1+3m^4 )^2 )/(4m^4 (1+m^2 )^2 )) ⇒ (((1−m^2 )^2 )/(1+3m^4 ))=((c^2 (1+3m^4 )^2 )/(4m^4 (1+m^2 )^2 )) ⇒ ((4m^4 (1−m^2 )^2 (1+m^2 )^2 )/((1+3m^4 )^3 ))=c^2 ⇒ ((4t(1−t)^2 )/((1+3t)^3 ))=c^2 _________________________
$${m}=\mathrm{tan}\:\theta=−\frac{{r}}{{k}} \\ $$$$\frac{\mathrm{2}{m}}{\mathrm{1}−{m}^{\mathrm{2}} }=\mathrm{tan}\:\mathrm{2}\theta=\frac{{p}−{r}}{{k}} \\ $$$$\left(−{km}\right)^{\mathrm{3}} =\left(−{km}\right)+{c}\:\:\:….\left({I}\right) \\ $$$$\frac{{p}}{{k}}=\frac{\mathrm{2}{m}−{m}+{m}^{\mathrm{3}} }{\mathrm{1}−{m}^{\mathrm{2}} }=\frac{{m}\left({m}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{1}−{m}^{\mathrm{2}} } \\ $$$${p}=\frac{{km}\left({m}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{1}−{m}^{\mathrm{2}} } \\ $$$${k}^{\mathrm{3}} {m}^{\mathrm{3}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} =\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} {km}\left({m}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\left({km}−{c}\right)\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} =\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} {km}\left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\mathrm{4}{km}^{\mathrm{3}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)={c}\left\{\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{3}} \right\} \\ $$$$\Rightarrow \\ $$$$\mathrm{4}{km}^{\mathrm{3}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)=\mathrm{2}{c}\left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right)\:\:…\left({II}\right) \\ $$$${from}\:\left({I}\right)\&\left({II}\right) \\ $$$$\mathrm{4}{km}^{\mathrm{3}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)=\mathrm{2}{km}\left(\mathrm{1}−{k}^{\mathrm{2}} {m}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:\:\mathrm{2}{m}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)=\left(\mathrm{1}−{k}^{\mathrm{2}} {m}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}^{\mathrm{4}} =\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} −{k}^{\mathrm{2}} {m}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:{k}^{\mathrm{2}} {m}^{\mathrm{2}} =\frac{{m}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} }\:=\:\frac{\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} } \\ $$$$\&\:\:\:{k}^{\mathrm{2}} {m}^{\mathrm{2}} =\frac{{c}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right)^{\mathrm{2}} }{\mathrm{4}{m}^{\mathrm{4}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} }=\frac{{c}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right)^{\mathrm{2}} }{\mathrm{4}{m}^{\mathrm{4}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{\mathrm{4}{m}^{\mathrm{4}} \left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{3}{m}^{\mathrm{4}} \right)^{\mathrm{3}} }={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{4}{t}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{3}{t}\right)^{\mathrm{3}} }={c}^{\mathrm{2}} \\ $$$$\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

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