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Question-174304




Question Number 174304 by dragan91 last updated on 29/Jul/22
Answered by mr W last updated on 29/Jul/22
x^5 −2x^4 −1=0  Σr=2  Σr_i r_j =0  (Σr)^2 =Σr^2 +2Σr_i r_j   2^2 =Σr^2 +2×0  ⇒Σr^2 =4    (1/x^8 )=(1/x^3 )−(2/x^4 )  (1/x^4 )=x−2  (1/x^3 )=x^2 −2x  Σ(1/r^8 )=Σ(1/r^3 )−2Σ(1/r^4 )  Σ(1/r^8 )=Σr^2 −2Σr−2(Σr−10)  Σ(1/r^8 )=Σr^2 −4Σr+20  Σ(1/r^8 )=4−4×2+20=16 ✓
$${x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\Sigma{r}=\mathrm{2} \\ $$$$\Sigma{r}_{{i}} {r}_{{j}} =\mathrm{0} \\ $$$$\left(\Sigma{r}\right)^{\mathrm{2}} =\Sigma{r}^{\mathrm{2}} +\mathrm{2}\Sigma{r}_{{i}} {r}_{{j}} \\ $$$$\mathrm{2}^{\mathrm{2}} =\Sigma{r}^{\mathrm{2}} +\mathrm{2}×\mathrm{0} \\ $$$$\Rightarrow\Sigma{r}^{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{2}}{{x}^{\mathrm{4}} } \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{4}} }={x}−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }={x}^{\mathrm{2}} −\mathrm{2}{x} \\ $$$$\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{8}} }=\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{3}} }−\mathrm{2}\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{4}} } \\ $$$$\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{8}} }=\Sigma{r}^{\mathrm{2}} −\mathrm{2}\Sigma{r}−\mathrm{2}\left(\Sigma{r}−\mathrm{10}\right) \\ $$$$\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{8}} }=\Sigma{r}^{\mathrm{2}} −\mathrm{4}\Sigma{r}+\mathrm{20} \\ $$$$\Sigma\frac{\mathrm{1}}{{r}^{\mathrm{8}} }=\mathrm{4}−\mathrm{4}×\mathrm{2}+\mathrm{20}=\mathrm{16}\:\checkmark \\ $$
Commented by Tawa11 last updated on 29/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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