Question

if m+1/m=-p, then show that m square +1/m square=p -2square​

Answer 1

Answer:

square it on both sides and using (a-b)^2 we will be able to proove.

Answer 2

Appropriate Question :-

$\rm \: If \: m + \dfrac{1}{m} =\: -\:p, \: show \: that \: {m}^{2} + \dfrac{1}{ {m}^{2} } = {p}^{2} - 2$

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

Given that,

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

On squaring both sides, we get

$\rm \: \bigg(m + \dfrac{1}{m}\bigg)^{2} = {p}^{2}$

We know,

$\boxed{\tt{ \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: }}$

So, using this identity, we get

$\rm \: {m}^{2} + {\bigg(\dfrac{1}{m} \bigg) }^{2} + 2 \times m \times \dfrac{1}{m} = {p}^{2}$

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

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

Hence, Proved

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More To Know :-

$\rm \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

$\rm \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2})$

$\rm \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$\rm \: {(a + b)}^{2} - 4ab = {(a - b)}^{2}$

$\rm \: {(a - b)}^{2} + 4ab = {(a + b)}^{2}$

$\rm \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$\rm \: (x + a)(x + b) = {x}^{2} + (a + b)x + ab$