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Question Number 190034 by Shrinava last updated on 26/Mar/23

$$\mathrm{If}\mathrm{x}+\frac{2}{\mathrm{x}}=\sqrt{17}$$$$\mathrm{Find}:\frac{({\mathrm{x}}^2-4)({\mathrm{x}}^2-1)}{{\mathrm{x}}^2}=?$$

Answered by BaliramKumar last updated on 26/Mar/23

$$x^2+\frac{4}{x^2}+2\cdot x\cdot \frac{2}{x}=17$$$$x^2+\frac{4}{x^2}+4=17$$$$x^4+4=13x^2.........\left(i\right)$$$$\frac{(x^2-4)(x^2-1)}{x^2}=\frac{x^4-5x^2+4}{x^2}=\frac{(x^4+4)-5x^2}{x^2}$$$$\frac{13x^2-5x^2}{x^2}=\frac{8x^2}{x^2}=8$$$$$$

Answered by Rasheed.Sindhi last updated on 26/Mar/23

$$\mathrm{If}\mathrm{x}+\frac{2}{\mathrm{x}}=\sqrt{17}$$$$\mathrm{Find}:\frac{({\mathrm{x}}^2-4)({\mathrm{x}}^2-1)}{{\mathrm{x}}^2}=?$$$${\mathrm{x}}^2=\sqrt{17}\mathrm{x}-2$$$$\frac{({\mathrm{x}}^2-4)({\mathrm{x}}^2-1)}{{\mathrm{x}}^2}$$$$=\frac{(\sqrt{17}\mathrm{x}-2-4)(\sqrt{17}\mathrm{x}-2-1)}{\sqrt{17}\mathrm{x}-2}$$$$=\frac{(\sqrt{17}\mathrm{x}-6)(\sqrt{17}\mathrm{x}-3)}{\sqrt{17}\mathrm{x}-2}$$$$=\frac{{17\mathrm{x}}^2-9\sqrt{17}\mathrm{x}+18}{\sqrt{17}\mathrm{x}-2}$$$$=\frac{17(\sqrt{17}\mathrm{x}-2)-9\sqrt{17}\mathrm{x}+18}{\sqrt{17}\mathrm{x}-2}$$$$=\frac{17\sqrt{17}\mathrm{x}-34-9\sqrt{17}\mathrm{x}+18}{\sqrt{17}\mathrm{x}-2}$$$$=\frac{8\sqrt{17}\mathrm{x}-16}{\sqrt{17}\mathrm{x}-2}=\frac{8(\sqrt{17}\mathrm{x}-2)}{\sqrt{17}\mathrm{x}-2}=8$$

Commented by Shrinava last updated on 01/Apr/23

$$\mathrm{cool}\mathrm{professor}\mathrm{thankyou}$$

Answered by Rasheed.Sindhi last updated on 26/Mar/23

$$\mathrm{x}+\frac{2}{\mathrm{x}}=\sqrt{17};\frac{({\mathrm{x}}^2-4)({\mathrm{x}}^2-1)}{{\mathrm{x}}^2}=?$$$$\frac{({\mathrm{x}}^2-4)({\mathrm{x}}^2-1)}{{\mathrm{x}}^2}=\frac{{\mathrm{x}}^2-4}{\mathrm{x}}\cdot \frac{{\mathrm{x}}^2-1}{\mathrm{x}}$$$$=(\mathrm{x}-\frac{4}{\mathrm{x}})(\mathrm{x}-\frac{1}{\mathrm{x}})$$$$={\mathrm{x}}^2+\frac{4}{{\mathrm{x}}^2}-5={(\mathrm{x}+\frac{2}{\mathrm{x}})}^2-4-5$$$$={\left(\sqrt{17}\right)}^2-9=17-9=8$$

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