Question

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

Answer 1

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

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

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

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

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

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 :-

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

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

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

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

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

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

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

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

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