p2+1p2=47,p2+1p2=47,{p^2} + \frac{1}{{{p^2}}} = 47{\text{,}} then the value of p+1p" role="presentation">p+1pp+1pp + \frac{1}{p} is?" /> p2+1p2=47,p2+1p2=47,{p^2} + \frac{1}{{{p^2}}} = 47{\text{,}} then the value of p+1p" role="presentation">p+1pp+1pp + \frac{1}{p} is?" /> p2+1p2=47,p2+1p2=47,{p^2} + \frac{1}{{{p^2}}} = 47{\text{,}} then the value of p+1p" role="presentation">p+1pp+1pp + \frac{1}{p} is?" />

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If $p^{2} + \frac{1}{p^{2}} = 47,$ ${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$ then the value of $p + \frac{1}{p}$ $p + \frac{1}{p}$ is?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} p^{2} + \frac{1}{p^{2}} = 47 \\ On adding 2 both side \\ p^{2} + \frac{1}{p^{2}} + 2 = 47 + 2 \\ \Rightarrow {(p + \frac{1}{p})}^{2} = 49 \\ \Rightarrow (p + \frac{1}{p}) = 7 \\ \Rightarrow p + \frac{1}{p} = 7 \end{aligned}

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then the value of

p + \frac{1}{p}

$p + \frac{1}{p}$ is?", "text": "If

p^{2} + \frac{1}{p^{2}} = 47,

${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$ then the value of

p + \frac{1}{p}

$p + \frac{1}{p}$ is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} p^{2} + \frac{1}{p^{2}} = 47 \\ On adding 2 both side \\ p^{2} + \frac{1}{p^{2}} + 2 = 47 + 2 \\ \Rightarrow {(p + \frac{1}{p})}^{2} = 49 \\ \Rightarrow (p + \frac{1}{p}) = 7 \\ \Rightarrow p + \frac{1}{p} = 7 \end{aligned}

$\eqalign{ & {p^2} + \frac{1}{{{p^2}}} = 47 \cr & {\text{On adding 2 both side}} \cr & {p^2} + \frac{1}{{{p^2}}} + 2 = 47 + 2 \cr & \Rightarrow {\left( {p + \frac{1}{p}} \right)^2} = 49 \cr & \Rightarrow \left( {p + \frac{1}{p}} \right) = 7 \cr & \Rightarrow p + \frac{1}{p} = 7 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: