Question

if p+1/p=2prove that p²+1/p²=p⁴+1/p⁴

Answer 1

Hey

Here is your answer,

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p+1 / p = 2

Cross multiplying

p+1 = 2p

2p-p = 1

p = 1

To prove : p²+1/p²=p⁴+1/p⁴

Substitute the value of p in the equation,

(1)² + 1 / (1)² = (1)⁴ + 1 / (1)⁴

1 + 1/ 1 = 1 + 1/1

2/1 = 2/1

2 = 2

LHS = RHS

Hence proved.

Hope it helps you!

Answer 2

$\mathsf{p + \dfrac{1}{p} = 2}$


$\mathsf{\texttt{ Square on both sides}}$


$\mathsf{ \implies \: \bigg(p + \dfrac{1}{p} \bigg) {}^{2} = 2 {}^{2} }$

$\mathsf{We\: know, ( a + b )^{2}= a^{2} + b^{2} + 2ab}$

$\therefore \mathsf{ {p}^{2} + \dfrac{1}{ {p}^{2} } + \bigg( 2 \times p \times \dfrac{1}{p } \bigg) } = 4 \\ \\ \\ \mathsf{ \implies \: {p}^{2} + \dfrac{1}{p {}^{2} } + 2(1) = 4} \\ \\ \\ \mathsf{ \implies \: {p}^{2} + \dfrac{1}{p {}^{2} } = 4 - 2} \\ \\ \\ \mathsf{ \implies p {}^{2} + \dfrac{1}{p {}^{2} } = 2} \: \: \: \: \: \: \: \: \: \bold{ \textit{...(i)}}$


$\mathsf{\texttt{ Square on both sides}}$


$\mathsf{ \implies \bigg(p {}^{2} + \dfrac{1}{p {}^{2} } \bigg) {}^{2} = 2 {}^{2} }$


$\mathsf{We\: know, ( a + b )^{2}= a^{2} + b^{2} + 2ab}$


$\therefore \mathsf{ {p}^{4} + \dfrac{1}{ {p}^{4} } + \bigg( 2 \times p {}^{4} \times \dfrac{1}{p {}^{4} } \bigg) } = 4 \\ \\ \\ \mathsf{ \implies \: {p}^{4} + \dfrac{1}{p {}^{4} } + 2(1) = 4} \\ \\ \\ \mathsf{ \implies \: {p}^{4} + \dfrac{1}{p {}^{4} } = 4 - 2} \\ \\ \\ \mathsf{ \implies p {}^{4} + \dfrac{1}{p {}^{4} } = 2} \: \: \: \: \: \: \: \: \: \bold{ \textit{...(ii)}}$



$\mathsf{From \bold{\textit{( i ) and ( ii )}, }2 = 2 }$


$\therefore \mathsf{ p {}^{2} + \dfrac{1}{ {p}^{2} } = {p}^{4} + \dfrac{1}{p {}^{4} }}$


Hence, proved.