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Question Number 134135 by Agnibhoo last updated on 28/Feb/21

$$\mathrm{If}\mathrm{P}=2+\frac{1}{\mathrm{P}}\mathrm{then}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{answer}\mathrm{of}$$$${\mathrm{P}}^2-\frac{1}{{\mathrm{P}}^2}?$$

Answered by Ñï= last updated on 28/Feb/21

$$p^2-\frac{1}{p^2}$$$$=(p-\frac{1}{p})(p+\frac{1}{p})$$$$=(p-\frac{1}{p}){[{(p-\frac{1}{p})}^2+4]}^{1/2}$$$$=2\times {(2^2+4)}^{1/2}$$$$=4\sqrt{2}$$

Answered by bobhans last updated on 28/Feb/21

$$p-\frac{1}{p}=2\Rightarrow p^2+\frac{1}{p^2}=6$$$$\Rightarrow {(p+\frac{1}{p})}^2-2=6\Rightarrow p+\frac{1}{p}=2\sqrt{2}$$$$then(p-\frac{1}{p})(p+\frac{1}{p})=2\times 2\sqrt{2}$$$$p^2-\frac{1}{p^2}=4\sqrt{2}$$

Answered by mr W last updated on 28/Feb/21

$$P-\frac{1}{P}=2$$$$P^2-2+\frac{1}{P^2}=4$$$$P^2+2+\frac{1}{P^2}=8$$$${(P+\frac{1}{P})}^2=8$$$$P+\frac{1}{P}=\pm 2\sqrt{2}$$$$P^2-\frac{1}{P^2}=(P+\frac{1}{P})(P-\frac{1}{P})=\pm 4\sqrt{2}$$

Answered by Rasheed.Sindhi last updated on 28/Feb/21

$$\mathrm{P}=2+\frac{1}{\mathrm{P}}\Rightarrow {\mathrm{p}}^2-2\mathrm{p}-1=0$$$$\mathrm{p}=1\pm \sqrt{2}\Rightarrow {\mathrm{p}}^2=1+2\pm 2\sqrt{2}=3\pm 2\sqrt{2}$$$${\mathrm{p}}^2-\frac{1}{{\mathrm{p}}^2}=(3\pm 2\sqrt{2})-\frac{1}{3\pm 2\sqrt{2}}.\frac{3\mp 2\sqrt{2}}{3\mp 2\sqrt{2}}$$$$=3\pm 2\sqrt{2}-\frac{3\mp 2\sqrt{2}}{9-8}=3\pm 2\sqrt{2}-3\pm 2\sqrt{2}$$$$=\pm 4\sqrt{2}$$

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