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Question Number 223964 by Rojarani last updated on 11/Aug/25
Answered by Rasheed.Sindhi last updated on 11/Aug/25
x^3 +2x^2 +3x+4=0 a,b,c are the roots a+b+c=−2 ab+bc+ca=3 abc=−4 (1+(2/(a−1)))(1+(2/(b−1)))(1+(2/(c−1))) =((a+1)/(a−1))∙((b+1)/(b−1))∙((c+1)/(c−1)) =((abc+(a+b+c)+(ab+bc+ca)+1)/(abc+(a+b+c)−(ab+bc+ca)−1)) =((−4+(−2)+3+1)/(−4+(−2)−3−1))=((−2)/(−10))=(1/5)
$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$${a},{b},{c}\:{are}\:{the}\:{roots} \\ $$$${a}+{b}+{c}=−\mathrm{2} \\ $$$${ab}+{bc}+{ca}=\mathrm{3} \\ $$$${abc}=−\mathrm{4} \\ $$$$\: \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2}}{{a}−\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{b}−\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{c}−\mathrm{1}}\right) \\ $$$$=\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}\centerdot\frac{{b}+\mathrm{1}}{{b}−\mathrm{1}}\centerdot\frac{{c}+\mathrm{1}}{{c}−\mathrm{1}} \\ $$$$=\frac{{abc}+\left({a}+{b}+{c}\right)+\left({ab}+{bc}+{ca}\right)+\mathrm{1}}{{abc}+\left({a}+{b}+{c}\right)−\left({ab}+{bc}+{ca}\right)−\mathrm{1}} \\ $$$$=\frac{−\mathrm{4}+\left(−\mathrm{2}\right)+\mathrm{3}+\mathrm{1}}{−\mathrm{4}+\left(−\mathrm{2}\right)−\mathrm{3}−\mathrm{1}}=\frac{−\mathrm{2}}{−\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by Rojarani last updated on 11/Aug/25
Sir, thanks.
$$\:{Sir},\:{thanks}. \\ $$
Answered by behi834171 last updated on 11/Aug/25
z^3 +2z^2 +3z+4=0 (z≠1) y=1+(2/(z−1))=((z+1)/(z−1)) ⇒(((z+1)/(z−1)))^3 +2(((z+1)/(z−1)))^2 +3(((z+1)/(z−1)))+4=0 ⇒(z+1)^3 + (z^3 +3z^2 +3z+1) +2(z+1)^2 (z−1)+ (2z^3 +2z^2 −2z−2) +3(z+1)(z−1)^2 + (3z^3 −3z^2 −3z+3) +4(z−1)^3 =0 (4z^3 −12z^2 +12z−4)⇒ ⇒10z^3 −10z^2 +10z−2=0⇒ ⇒z^3 −z^2 +z−(1/5)=0⇒ ⇒Π(1+(2/(a−1)))=Π_1 ^3 z_i =(1/5) .■
$$\boldsymbol{{z}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{z}}+\mathrm{4}=\mathrm{0}\:\:\:\:\left(\boldsymbol{{z}}\neq\mathrm{1}\right) \\ $$$$\boldsymbol{{y}}=\mathrm{1}+\frac{\mathrm{2}}{\boldsymbol{{z}}−\mathrm{1}}=\frac{\boldsymbol{{z}}+\mathrm{1}}{\boldsymbol{{z}}−\mathrm{1}} \\ $$$$\Rightarrow\left(\frac{\boldsymbol{{z}}+\mathrm{1}}{\boldsymbol{{z}}−\mathrm{1}}\right)^{\mathrm{3}} +\mathrm{2}\left(\frac{\boldsymbol{{z}}+\mathrm{1}}{\boldsymbol{{z}}−\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{\boldsymbol{{z}}+\mathrm{1}}{\boldsymbol{{z}}−\mathrm{1}}\right)+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\left(\boldsymbol{{z}}+\mathrm{1}\right)^{\mathrm{3}} + \\ $$$$\left(\boldsymbol{{z}}^{\mathrm{3}} +\mathrm{3}\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{z}}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(\boldsymbol{{z}}+\mathrm{1}\right)^{\mathrm{2}} \left(\boldsymbol{{z}}−\mathrm{1}\right)+ \\ $$$$\left(\mathrm{2}\boldsymbol{{z}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{z}}−\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\left(\boldsymbol{{z}}+\mathrm{1}\right)\left(\boldsymbol{{z}}−\mathrm{1}\right)^{\mathrm{2}} + \\ $$$$\left(\mathrm{3}\boldsymbol{{z}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{z}}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\boldsymbol{{z}}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mathrm{4}\boldsymbol{{z}}^{\mathrm{3}} −\mathrm{12}\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{12}\boldsymbol{{z}}−\mathrm{4}\right)\Rightarrow \\ $$$$\Rightarrow\mathrm{10}\boldsymbol{{z}}^{\mathrm{3}} −\mathrm{10}\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{{z}}−\mathrm{2}=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{z}}^{\mathrm{3}} −\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{z}}−\frac{\mathrm{1}}{\mathrm{5}}=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow\Pi\left(\mathrm{1}+\frac{\mathrm{2}}{\boldsymbol{{a}}−\mathrm{1}}\right)=\underset{\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\boldsymbol{{z}}_{\boldsymbol{{i}}} =\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:.\blacksquare \\ $$$$ \\ $$
Answered by mr W last updated on 12/Aug/25
x^3 +2x^2 +3x+4=0 (x−1+1)^3 +2(x−1+1)^2 +3(x−1+1)+4=0 (u+1)^3 +2(u+1)^2 +3(u+1)+4=0 u^3 +3u^2 +3u+1+2u^2 +4u+2+3u+3+4=0 u^3 +5u^2 +10u+10=0 ((2/u)+1−1)^3 +2((2/u)+1−1)^2 +2((2/u)+1−1)+(4/5)=0 (v−1)^3 +2(v−1)^2 +2(v−1)+(4/5)=0 ...−1+2−2+(4/5)=0 ...−(1/5)=0 v=(2/u)+1=1+(2/(x−1)) (1+(2/(a−1)))(1+(2/(b−1)))(1+(2/(c−1))) =v_1 v_2 v_3 =−(−(1/5))=(1/5) ✓
$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({x}−\mathrm{1}+\mathrm{1}\right)+\mathrm{4}=\mathrm{0} \\ $$$$\left({u}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({u}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({u}+\mathrm{1}\right)+\mathrm{4}=\mathrm{0} \\ $$$${u}^{\mathrm{3}} +\mathrm{3}{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +\mathrm{4}{u}+\mathrm{2}+\mathrm{3}{u}+\mathrm{3}+\mathrm{4}=\mathrm{0} \\ $$$${u}^{\mathrm{3}} +\mathrm{5}{u}^{\mathrm{2}} +\mathrm{10}{u}+\mathrm{10}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}}{{u}}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left(\frac{\mathrm{2}}{{u}}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{2}}{{u}}+\mathrm{1}−\mathrm{1}\right)+\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0} \\ $$$$\left({v}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({v}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left({v}−\mathrm{1}\right)+\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0} \\ $$$$…−\mathrm{1}+\mathrm{2}−\mathrm{2}+\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0} \\ $$$$…−\frac{\mathrm{1}}{\mathrm{5}}=\mathrm{0} \\ $$$${v}=\frac{\mathrm{2}}{{u}}+\mathrm{1}=\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2}}{{a}−\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{b}−\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{c}−\mathrm{1}}\right) \\ $$$$\:\:={v}_{\mathrm{1}} {v}_{\mathrm{2}} {v}_{\mathrm{3}} =−\left(−\frac{\mathrm{1}}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{5}}\:\checkmark \\ $$

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