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x-3-4x-2-9x-14-0-1-1-x-1-2-1-x-2-2-1-x-3-2-2-x-1-2-x-2-2-x-3-

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Question Number 174356 by behi834171 last updated on 30/Jul/22
x^3 +4x^2 −9x−14=0 1. (1/x_1 ^2 )+(1/x_2 ^2 )+(1/x_3 ^2 )=? 2. Σ ((x_1 ^2 +x_2 ^2 )/x_3 )=?
$$\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\boldsymbol{{x}}−\mathrm{14}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{1}.\:\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{3}} ^{\mathrm{2}} }=? \\ $$$$\:\:\:\:\:\mathrm{2}.\:\Sigma\:\frac{\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }{\boldsymbol{{x}}_{\mathrm{3}} }=? \\ $$
Answered by behi834171 last updated on 30/Jul/22
let: y=(1/x)⇒(1/y^3 )+(4/y^2 )−(9/y)−14=0 ⇒y^3 +(9/(14))y^2 −(2/7)y−(1/(14))=0 Σ(1/x_i ^2 )=Σy_i ^2 =(Σy)^2 −2(Σy_i .y_j )= =(−(9/(14)))^2 −2(−(2/7))=((81)/(196))+(4/7)=((81+112)/(196))= =((193)/(196)) .■ ((a^2 +b^2 )/c)=(((a+b)^2 −2ab)/c)=(((−4−c)^2 −2((abc)/c))/c)= =((16+8c+c^2 −((28)/c))/c)=((16c+8c^2 +c^3 −28)/c^2 )= =((16)/c)+c−((28)/c^2 )+8⇒ Σ((x_1 ^2 +x_2 ^2 )/x_3 )=24+Σx_i +16Σ(1/x_i )−28Σ(1/x_i ^2 )= =24+(−4)+16(−(9/(14)))−28(((193)/(196)))= =((−28×193+24×196−14×16×9−4×196)/(196))= =−((875)/(49)) . ■
$${let}:\:\:\:{y}=\frac{\mathrm{1}}{{x}}\Rightarrow\frac{\mathrm{1}}{{y}^{\mathrm{3}} }+\frac{\mathrm{4}}{{y}^{\mathrm{2}} }−\frac{\mathrm{9}}{{y}}−\mathrm{14}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{y}}^{\mathrm{3}} +\frac{\mathrm{9}}{\mathrm{14}}\boldsymbol{{y}}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{7}}\boldsymbol{{y}}−\frac{\mathrm{1}}{\mathrm{14}}=\mathrm{0} \\ $$$$\Sigma\frac{\mathrm{1}}{\boldsymbol{{x}}_{\boldsymbol{{i}}} ^{\mathrm{2}} }=\Sigma\boldsymbol{{y}}_{\boldsymbol{{i}}} ^{\mathrm{2}} =\left(\Sigma\boldsymbol{{y}}\right)^{\mathrm{2}} −\mathrm{2}\left(\Sigma\boldsymbol{{y}}_{\boldsymbol{{i}}} .\boldsymbol{{y}}_{\boldsymbol{{j}}} \right)= \\ $$$$=\left(−\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} −\mathrm{2}\left(−\frac{\mathrm{2}}{\mathrm{7}}\right)=\frac{\mathrm{81}}{\mathrm{196}}+\frac{\mathrm{4}}{\mathrm{7}}=\frac{\mathrm{81}+\mathrm{112}}{\mathrm{196}}= \\ $$$$=\frac{\mathrm{193}}{\mathrm{196}}\:\:\:\:.\blacksquare \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}}=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}}{{c}}=\frac{\left(−\mathrm{4}−{c}\right)^{\mathrm{2}} −\mathrm{2}\frac{{abc}}{{c}}}{{c}}= \\ $$$$=\frac{\mathrm{16}+\mathrm{8}{c}+{c}^{\mathrm{2}} −\frac{\mathrm{28}}{{c}}}{{c}}=\frac{\mathrm{16}{c}+\mathrm{8}{c}^{\mathrm{2}} +{c}^{\mathrm{3}} −\mathrm{28}}{{c}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{16}}{\boldsymbol{{c}}}+\boldsymbol{{c}}−\frac{\mathrm{28}}{\boldsymbol{{c}}^{\mathrm{2}} }+\mathrm{8}\Rightarrow \\ $$$$\Sigma\frac{\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }{\boldsymbol{{x}}_{\mathrm{3}} }=\mathrm{24}+\Sigma\boldsymbol{{x}}_{\boldsymbol{{i}}} +\mathrm{16}\Sigma\frac{\mathrm{1}}{\boldsymbol{{x}}_{\boldsymbol{{i}}} }−\mathrm{28}\Sigma\frac{\mathrm{1}}{\boldsymbol{{x}}_{\boldsymbol{{i}}} ^{\mathrm{2}} }= \\ $$$$=\mathrm{24}+\left(−\mathrm{4}\right)+\mathrm{16}\left(−\frac{\mathrm{9}}{\mathrm{14}}\right)−\mathrm{28}\left(\frac{\mathrm{193}}{\mathrm{196}}\right)= \\ $$$$=\frac{−\mathrm{28}×\mathrm{193}+\mathrm{24}×\mathrm{196}−\mathrm{14}×\mathrm{16}×\mathrm{9}−\mathrm{4}×\mathrm{196}}{\mathrm{196}}= \\ $$$$=−\frac{\mathrm{875}}{\mathrm{49}}\:\:\:\:\:.\:\blacksquare \\ $$
Commented by Tawa11 last updated on 31/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by BaliramKumar last updated on 30/Jul/22
1. (1/α^2 ) + (1/β^2 ) + (1/γ^2 ) = ((1/α)+(1/β)+(1/γ))^2 − 2((1/(αβ)) + (1/(βγ)) + (1/(γα))) = (((αβ+βγ+γα)/(αβγ)))^2 − 2(((α+β+γ)/(αβγ))) = (((−9)/(14)))^2 −2(((−4)/(14))) = ((81)/(196)) + (8/(14)) = ((193)/(196))
$$\mathrm{1}.\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\gamma^{\mathrm{2}} }\:=\:\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left(\frac{\mathrm{1}}{\alpha\beta}\:+\:\frac{\mathrm{1}}{\beta\gamma}\:+\:\frac{\mathrm{1}}{\gamma\alpha}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left(\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{−\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\frac{−\mathrm{4}}{\mathrm{14}}\right)\:=\:\frac{\mathrm{81}}{\mathrm{196}}\:+\:\frac{\mathrm{8}}{\mathrm{14}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{193}}{\mathrm{196}}\: \\ $$
Commented by behi834171 last updated on 30/Jul/22
thank you sir for your time. right answer✓.
$${thank}\:{you}\:{sir}\:{for}\:{your}\:{time}. \\ $$$${right}\:{answer}\checkmark. \\ $$
Commented by Tawa11 last updated on 31/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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