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Question Number 205073 by mr W last updated on 07/Mar/24

$$ifa,b,caretherootsof$$$$f\left(x\right)=x^3-2024x^2+2024x+2024$$$$find\frac{1}{1-a^2}+\frac{1}{1-b^2}+\frac{1}{1-c^2}=?$$

Answered by cortano12 last updated on 07/Mar/24

$$=\frac{{\mathrm{a}}^2}{1-{\mathrm{a}}^2}+\frac{{\mathrm{b}}^2}{1-{\mathrm{b}}^2}+\frac{{\mathrm{c}}^2}{1-{\mathrm{c}}^2}+3$$$$=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-2{\left(\mathrm{ab}\right)}^2}{(1-{\mathrm{a}}^2)(1-{\mathrm{b}}^2)}+\frac{{\mathrm{c}}^2}{1-{\mathrm{c}}^2}+3$$$$=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+3{\left(\mathrm{abc}\right)}^2-2({\left(\mathrm{ab}\right)}^2+{\left(\mathrm{ac}\right)}^2+{\left(\mathrm{bc}\right)}^2}{(1-{\mathrm{a}}^2)(1-{\mathrm{b}}^2)(1-{\mathrm{c}}^2)}+3$$

Answered by Frix last updated on 07/Mar/24

$$y=\frac{1}{1-x^2}\Rightarrow x=\pm \sqrt{\frac{y-1}{y}}$$$$\mathrm{Inserting}\mathrm{\&}\mathrm{transforming}$$$$y^3-\frac{2027y^2}{2025}-\frac{2021y}{2025}-\frac{1}{4100625}=0$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{2027}{2025}$$

Answered by A5T last updated on 07/Mar/24

$$WLOG,leta=x+yi,b=x-yi,c=cwherex,y,c\in \mathbb{R}$$$$?=\frac{1}{1-{(x+yi)}^2}+\frac{1}{1-{(x-yi)}^2}+\frac{1}{1-c^2}$$$$\Rightarrow 2?=\frac{1}{1-x-yi}+\frac{1}{1-x+yi}+\frac{1}{1-c}+\frac{1}{1+x+yi}+\frac{1}{1+x-yi}+\frac{1}{1+c}$$$$=\frac{2(1-x)}{{(1-x)}^2+y^2}+\frac{1}{1-c}+\frac{2(1+x)}{{(1+x)}^2+y^2}+\frac{1}{1+c}$$$$=\frac{2(1-c-x+xc)+1+x^2-2x+y^2}{1+x^2-2x+y^2-c-{cx}^2+2cx-{cy}^2}+\frac{2(1+x+c+cx)+1+x^2+2x+y^2}{1+x^2+2x+y^2+c+{cx}^2+2cx+{cy}^2}$$$$=\frac{3-2(c+2x)+x^2+y^2+2cx}{1+x^2+y^2+2cx-(2x+c)-c(x^2+y^2)}+\frac{3+x^2+y^2+2cx+2(2x+c)}{1+x^2+y^2+2cx+2x+c+c(x^2+y^2)}$$$$=\frac{3-\left(2024\right)}{2025}+\frac{3\times 2025}{2025}=\frac{4054}{2025}=2?\Rightarrow ?=\frac{2027}{2025}$$$$[ab+bc+ca=2024\Rightarrow x^2+y^2+2cx=2024$$$$abc=(x^2+y^2)c=-2024;a+b+c=2x+c=2024]$$

Answered by A5T last updated on 07/Mar/24

$$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2?$$$$\frac{\mathrm{\Sigma }(1-a)(1-b)}{(1-a)(1-b)(1-c)}+\frac{\mathrm{\Sigma }(1+a)(1+b)}{(1+a)(1+b)(1+c)}$$$$=\frac{3-2(a+b+c)+ab+bc+ca}{1-a-b-c+ab+ac+bc-abc}+\frac{3+2\left(\mathrm{\Sigma }a\right)+\mathrm{\Sigma }ab}{1+\mathrm{\Sigma }a+\mathrm{\Sigma }ab+abc}$$$$=\frac{3-2\left(2024\right)+2024}{1-2024+2024+2024}+\frac{3+2\left(2024\right)+2024}{1+2024+2024-2024}$$$$=\frac{3-2024}{2025}+\frac{3\times 2025}{2025}=\frac{4054}{2025}=2?\Rightarrow ?=\frac{2027}{2025}$$

Answered by mr W last updated on 07/Mar/24

$${(x-1+1)}^3-2024{(x-1+1)}^2+2024(x-1+1)+2024=0$$$$lett=x-1$$$${(t+1)}^3-2024{(t+1)}^2+2024(t+1)+2024=0$$$$\frac{2025}{t^3}-\frac{2021}{t^2}-\frac{2021}{t}+1=0$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{1-x}=-\mathrm{\Sigma }\frac{1}{t}=-\frac{2021}{2025}$$$${(x+1-1)}^3-2024{(x+1-1)}^2+2024(x+1-1)+2024=0$$$$lets=x+1$$$${(s-1)}^3-2024{(s-1)}^2+2024(s-1)+2024=0$$$$s^3-2027s^2+6075s-2025=0$$$$\frac{2025}{s^3}-\frac{6075}{s^2}+\frac{2027}{s}-1=0$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{1+x}=\mathrm{\Sigma }\frac{1}{s}=\frac{6075}{2025}$$$$\frac{1}{1-a^2}+\frac{1}{1-b^2}+\frac{1}{1-c^2}$$$$=\frac{1}{2}(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c})$$$$=\frac{1}{2}(\mathrm{\Sigma }\frac{1}{1-x}+\mathrm{\Sigma }\frac{1}{1+x})$$$$=\frac{1}{2}(-\frac{2021}{2025}+\frac{6075}{2025})=\frac{2027}{2025}✓$$

Answered by pi314 last updated on 08/Mar/24

$$\frac{P^{\prime }\left(x\right)}{P\left(x\right)}=\sum _{a,b,c}\frac{1}{x-c};\frac{3x^2-4048x+2024}{x^3-2024x^2+2024x+2024}$$$$S=\frac{1}{2}(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c})+\frac{1}{2}(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c})$$$$S=\frac{1}{2}(\frac{P^{\prime }\left(1\right)}{P\left(1\right)}-\frac{P^{\prime }(-1)}{P(-1)})=\frac{1}{2}(\frac{-2021}{2025}+\frac{6075}{2025})=\frac{2027}{2025}$$

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