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Question Number 145114 by imjagoll last updated on 02/Jul/21

Commented by MJS_new last updated on 02/Jul/21

$$\mathrm{for}x\in \mathbb{R}\mathrm{I}\mathrm{get}2+\sqrt{5}$$

Commented by imjagoll last updated on 02/Jul/21

$$\mathrm{how}\mathrm{did}\mathrm{the}\mathrm{answer}\mathrm{sir}?$$

Commented by Canebulok last updated on 02/Jul/21

$$\mathit{R}\mathit{a}\mathit{n}\mathit{d}\mathit{o}\mathit{m}\mathit{P}\mathit{r}\mathit{o}\mathit{b}\mathit{l}\mathit{e}\mathit{m}:$$$$If{x}^4+x^2=\frac{11}{5}$$$$Whatisthevalueof$$$${}^3\sqrt{\frac{x+1}{x-1}}+{}^3\sqrt{\frac{x-1}{x+1}}=?$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$\Rightarrow 5x^4+5x^2-11=0$$$$let:$$$$\Rightarrow x^2=K$$$$\because$$$$\Rightarrow 5K^2+5K-11=0$$$$Byquadraticformula,$$$$\Rightarrow K=\frac{-5\pm \sqrt{5^2-4\left(5\right)(-11)}}{2\left(5\right)}$$$$\Rightarrow K_1=\frac{-5+7\sqrt{5}}{10}$$$$\Rightarrow K_2=\frac{-5-7\sqrt{5}}{10}$$$$\therefore$$$$\Rightarrow x^2=\frac{-5\pm 7\sqrt{5}}{10}$$$$\Rightarrow x=\sqrt{\left(\frac{-5\pm 7\sqrt{5}}{10}\right)}$$$$Thus;$$$$\Rightarrow {}^3\sqrt{\frac{x+1}{x-1}}+{}^3\sqrt{\frac{x-1}{x+1}}\approx 1.837977$$$$or$$$$\Rightarrow {}^3\sqrt{\frac{x+1}{x-1}}+{}^3\sqrt{\frac{x-1}{x+1}}\approx 4.236068$$$$$$

Commented by Canebulok last updated on 02/Jul/21

sorry i didn't notice

Answered by liberty last updated on 02/Jul/21

$$\Rightarrow 5x^4+5x^2-11=0$$$$\Rightarrow x^2=\frac{-5+\sqrt{25+220}}{10}$$$$\Rightarrow x^2+1=\frac{5+\sqrt{245}}{10}$$$$\Rightarrow x^2+1=\frac{5+7\sqrt{5}}{10}...\left(i\right)$$$$\Rightarrow x^2-1=\frac{-15+7\sqrt{5}}{10}...\left(ii\right)$$$$\Rightarrow \frac{x^2+1}{x^2-1}=\frac{5+7\sqrt{5}}{-15+7\sqrt{5}}...\left(iii\right)$$$$consider\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$$$$=\frac{\frac{x+1}{x-1}+\frac{x-1}{x+1}}{\sqrt[3]{{\left(\frac{x+1}{x-1}\right)}^2}+\sqrt[3]{{\left(\frac{x-1}{x+1}\right)}^2}-1}$$$$=\frac{\frac{2(x^2+1)}{x^2-1}}{\sqrt[3]{\frac{{(x+1)}^3}{{(x^2-1)}^2}}+\sqrt[3]{\frac{{(x-1)}^3}{{(x^2-1)}^2}}-1}$$$$=\frac{2\left(\frac{7\sqrt{5}+5}{7\sqrt{5}-15}\right)}{\frac{2x}{\sqrt[3]{{\left(\frac{7\sqrt{5}-15}{10}\right)}^2}}-1}$$$$...$$

Answered by MJS_new last updated on 02/Jul/21

$$a^{1/3}+b^{1/3}=c$$$$a+3a^{1/3}{b}^{1/3}(a^{1/3}+b^{1/3})+b=c^3$$$$a+3a^{1/3}{b}^{1/3}c+b=c^3$$$$3a^{1/3}{b}^{1/3}c=c^3-a-b$$$$27{abc}^3={(c^3-a-b)}^3$$$$c^9-3(a+b)c^6+3(a^2-7ab+b^2)c^3-{(a+b)}^3=0$$$$a=\frac{x+1}{x-1}\wedge b=\frac{1}{a}$$$$c^9-\frac{6(x^2+1)}{x^2-1}{c}^6-\frac{3(x^2-5)(5x^2-1)}{{(x^2-1)}^2}{c}^3-\frac{8{(x^2+1)}^3}{{(x^2-1)}^3}=0$$$$x^4+x^2=\frac{11}{5}\Rightarrow x^2=-\frac{1}{2}+\frac{7\sqrt{5}}{10}$$$$c^9-6(16+7\sqrt{5})c^6+21(285+128\sqrt{5})c^3-8(15856+7091\sqrt{5})=0$$$$c={(z+2(16+7\sqrt{5}))}^{1/3}$$$$z^3-27z-54(16+7\sqrt{5})=0$$$$(z-3(2+\sqrt{5}))(z^2+3(2+\sqrt{5})z+18(3+2\sqrt{5}))=0$$$$\Rightarrow$$$$z=3(2+\sqrt{5})\Rightarrow c=\sqrt[3]{38+17\sqrt{5}}=2+\sqrt{5}$$

Commented by MJS_new last updated on 02/Jul/21

$$\mathrm{is}\mathrm{this}\mathrm{really}\mathrm{so}\mathrm{hard}?\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{believe}...$$

Commented by imjagoll last updated on 03/Jul/21

$$\mathrm{hahaha}\mathrm{i}\mathrm{believe}\mathrm{it}\mathrm{problem}\mathrm{super}$$$$\mathrm{hard}\mathrm{sir}$$

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