Question

If p=5-2√6 ; find p to the power 2 +1/p to the power 2

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\underline{ \sf Given} \\ \star \sf \quad p = 5 - 2 \sqrt{6} \\ \\ \underline{ \sf To \: Find} \\ \star \quad \sf {p}^{2} + \frac{1}{ {p}^{2} } \\ \\ \underline{ \sf Solution} \\ \sf \rightarrow \quad p = 5 - 2 \sqrt{6} \\ \\ \sf \rightarrow \quad \frac{1}{p} = \frac{1}{5 - 2 \sqrt{6} } \\ \\ \sf \rightarrow \quad \frac{1}{p} = \frac{1}{5 - 2 \sqrt{6} } \times \frac{5 + 2 \sqrt{6} }{5 + 2 \sqrt{6} } \\ \\ \sf \rightarrow \quad \frac{1}{p} = \frac{5 + 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6} )}^{2} } \\ \\ \sf \rightarrow \quad \frac{1}{p} = \frac{5 + 2 \sqrt{6} }{25 - 24} \\ \\ \sf \rightarrow \quad \frac{1}{p} = 5 + 2 \sqrt{6} \\ \\ \quad \sf Add \: p \: both \: sides \\ \\ \sf \rightarrow \quad p + \frac{1}{p} = 5 + \cancel{2 \sqrt{6}} + 5 - \cancel{ 2 \sqrt{6}} \\ \\ \sf \rightarrow \quad p + \frac{1}{p} = 10 \\ \\ \sf \quad Square \: both \: sides \\ \\ \sf \rightarrow \quad \bigg(p + \frac{1}{p} \bigg)^{2} = {10}^{2} \\ \\ \sf \rightarrow \quad {p}^{2} + \frac{1}{ {p}^{2} } + 2 \times \cancel{p} \times \frac{1}{ \cancel{p}} = 100 \\ \\ \sf \rightarrow \quad {p}^{2} + \frac{1}{ {p}^{2} } = 100 - 2 \\ \\ \sf \rightarrow \quad \boxed{\sf \red{{p}^{2} + \frac{1}{ {p}^{2} } = 98}} \\ \\ \underline{ \sf Properties \: Used } \\ \\ \sf \hookrightarrow \quad {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$