Math, asked by anjalagurung317, 1 day ago

if m+1/m =-p then show that m2+1/m2=p2-2​

Answers

Answered by mathdude500
15

\large\underline{\sf{Given- }}

\rm :\longmapsto\:m +  \dfrac{1}{m}  =  \:  -  \: p

\large\underline{\sf{To\:prove \: - }}

\rm :\longmapsto\: {m}^{2}  +  \dfrac{1}{ {m}^{2} } =  {p}^{2}  - 2

\large\underline{\sf{Solution-}}

Given that

\rm :\longmapsto\:m +  \dfrac{1}{m}  =  \:  -  \: p

On squaring both sides, we get

\rm :\longmapsto\: {\bigg[ m +  \dfrac{1}{ m} \bigg]}^{2} =  {( - p)}^{2}

\rm :\longmapsto\: {m}^{2} + \dfrac{1}{ {m}^{2} } + 2 \times m \times \dfrac{1}{m} =  {p}^{2}

\rm :\longmapsto\: {m}^{2} + \dfrac{1}{ {m}^{2} } + 2  =  {p}^{2}

\rm :\longmapsto\:\boxed{ \tt{ \:  {m}^{2} + \dfrac{1}{ {m}^{2} } =  {p}^{2} - 2 \: }}

Hence, Proved

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More Identities to know :-

\boxed{ \tt{ \:  {(x - y)}^{2} =  {x}^{2}  +  {y}^{2}  - 2xy \: }}

\boxed{ \tt{ \:  {(x  +  y)}^{2} =  {x}^{2}  +  {y}^{2} + 2xy \: }}

\boxed{ \tt{ \:  {(x + y)}^{2} -  {(x - y)}^{2}  = 4xy \: }}

\boxed{ \tt{ \:  {(x + y)}^{2}  +  {(x - y)}^{2}  = 2( {x}^{2} +  {y}^{2}  \: }}

\boxed{ \tt{ \:  {x}^{2} -  {y}^{2}  = (x + y)(x - y)}}

\boxed{ \tt{ \:  {(x + y)}^{3}  =  {x}^{3} +  {y}^{3} + 3xy(x + y) \: }}

\boxed{ \tt{ \:  {(x + y)}^{3}  =  {x}^{3} + {3x}^{2}y + 3 {xy}^{2}  +{y}^{3} \: }}

\boxed{ \tt{ \:  {(x - y)}^{3}  =  {x}^{3} -  {3x}^{2}y + 3 {xy}^{2} - {y}^{3} \: }}

\boxed{ \tt{ \:  {(x - y)}^{3}  =  {x}^{3}-{y}^{3}  - 3xy(x - y) \: }}

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