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Question Number 163075 by tounghoungko last updated on 03/Jan/22

$$\frac{x^3+1}{x^2-1}=x+\sqrt{\frac{6}{x}}$$

Answered by Rasheed.Sindhi last updated on 03/Jan/22

$$\frac{x^3+1}{x^2-1}=x+\sqrt{\frac{6}{x}}$$$$\frac{x^3+1}{x^2-1}-x=\sqrt{\frac{6}{x}}$$$${\left(\frac{x^3+1-x^3+x}{x^2-1}\right)}^2=\frac{6}{x}$$$${\left(\frac{x+1}{(x-1)(x+1)}\right)}^2=\frac{6}{x}$$$${\left(\frac{1}{x-1}\right)}^2=\frac{6}{x}$$$$6(x^2-2x+1)=x$$$$6x^2-13x+6=0$$$$(3x-2)(2x-3)=0$$$$x=\frac{2}{3}\mid x=\frac{3}{2}$$$$Oncheckingitcanbeshownthat$$$$x=\frac{3}{2}isvalidroot,whilex=\frac{2}{3}is$$$$anextraneousroot.$$$$\therefore x=\frac{3}{2}$$

Answered by Rasheed.Sindhi last updated on 03/Jan/22

$$\frac{x^3+1}{x^2-1}=x+\sqrt{\frac{6}{x}}$$$$\frac{\cancel{(x+1)}(x^2-x+1}{\cancel{(x+1)}(x-1)}-x=\sqrt{\frac{6}{x}}$$$${\left(\frac{\cancel{x^2}-\cancel{x}+1-\cancel{x^2}+\cancel{x}}{x-1}\right)}^2=\frac{6}{x}$$$$6(x^2-2x+1)=x$$$$6x^2-13x+6=0$$$$(3x-2)(2x-3)=0$$$$\underset{\underset{\underset{\mathcal{R}oot}{\mathcal{E}xtraneous}}{}}{x=\frac{2}{3}}\mid x=\frac{3}{2}✓$$$$x=\frac{3}{2}$$

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