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Question Number 174356 by behi834171 last updated on 30/Jul/22

$${\mathit{x}}^3+4{\mathit{x}}^2-9\mathit{x}-14=0$$$$1.\frac{1}{{\mathit{x}}_1^2}+\frac{1}{{\mathit{x}}_2^2}+\frac{1}{{\mathit{x}}_3^2}=?$$$$2.\mathrm{\Sigma }\frac{{\mathit{x}}_1^2+{\mathit{x}}_2^2}{{\mathit{x}}_3}=?$$

Answered by behi834171 last updated on 30/Jul/22

$$let:y=\frac{1}{x}\Rightarrow \frac{1}{y^3}+\frac{4}{y^2}-\frac{9}{y}-14=0$$$$\Rightarrow {\mathit{y}}^3+\frac{9}{14}{\mathit{y}}^2-\frac{2}{7}\mathit{y}-\frac{1}{14}=0$$$$\mathrm{\Sigma }\frac{1}{{\mathit{x}}_{\mathit{i}}^2}=\mathrm{\Sigma }{\mathit{y}}_{\mathit{i}}^2={\left(\mathrm{\Sigma }\mathit{y}\right)}^2-2(\mathrm{\Sigma }{\mathit{y}}_{\mathit{i}}.{\mathit{y}}_{\mathit{j}})=$$$$={(-\frac{9}{14})}^2-2(-\frac{2}{7})=\frac{81}{196}+\frac{4}{7}=\frac{81+112}{196}=$$$$=\frac{193}{196}.◼$$$$\frac{a^2+b^2}{c}=\frac{{(a+b)}^2-2ab}{c}=\frac{{(-4-c)}^2-2\frac{abc}{c}}{c}=$$$$=\frac{16+8c+c^2-\frac{28}{c}}{c}=\frac{16c+8c^2+c^3-28}{c^2}=$$$$=\frac{16}{\mathit{c}}+\mathit{c}-\frac{28}{{\mathit{c}}^2}+8\Rightarrow$$$$\mathrm{\Sigma }\frac{{\mathit{x}}_1^2+{\mathit{x}}_2^2}{{\mathit{x}}_3}=24+\mathrm{\Sigma }{\mathit{x}}_{\mathit{i}}+16\mathrm{\Sigma }\frac{1}{{\mathit{x}}_{\mathit{i}}}-28\mathrm{\Sigma }\frac{1}{{\mathit{x}}_{\mathit{i}}^2}=$$$$=24+(-4)+16(-\frac{9}{14})-28\left(\frac{193}{196}\right)=$$$$=\frac{-28\times 193+24\times 196-14\times 16\times 9-4\times 196}{196}=$$$$=-\frac{875}{49}.◼$$

Commented by Tawa11 last updated on 31/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by BaliramKumar last updated on 30/Jul/22

$$1.\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}={(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^2-2(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })$$$$={\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)}^2-2\left(\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }\right)$$$$={\left(\frac{-9}{14}\right)}^2-2\left(\frac{-4}{14}\right)=\frac{81}{196}+\frac{8}{14}$$$$=\frac{193}{196}$$

Commented by behi834171 last updated on 30/Jul/22

$$thankyousirforyourtime.$$$$rightanswer✓.$$

Commented by Tawa11 last updated on 31/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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