Question

If m-1/m = p-2 then, prove that m²+1/m² = p²-4p+6 ​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $m-\frac{1}{m}=p-2$, then prove that $m^{2}+\frac{1}{m^{2}}=p^{2}-4p+6$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

Given that,

$m - \frac{1}{m} = p - 2$

Squaring both side, we get,

$(m - \frac{1}{m})^{2} =( p - 2)^{2}$

$\implies {m}^{2} + \frac{1}{ {m}^{2} } - 2 \times \cancel{m} \times \frac{1}{ \cancel{m}} = {p}^{2} - 2 \times p \times 2 + {2}^{2}$

$\implies {m}^{2} + \frac{1}{ {m}^{2} } - 2= {p}^{2} - 4p + 4$

$\implies {m}^{2} + \frac{1}{ {m}^{2} } = {p}^{2} - 4p + 4 + 2$

$\implies {m}^{2} + \frac{1}{ {m}^{2} } = {p}^{2} - 4p +6$

$\star\:\:\:\bf\large\underline\blue{Hence\:Proved.}$

Answer 2

$Given \: m - \frac{1}{m} = p - 2 \: ---(1)$

/* On squaring both sides of the equation,we get*/

$\implies \Big(m - \frac{1}{m} \Big)^{2} = ( p - 2)^{2}$

$\implies m^{2} + \frac{1}{m^{2}} - 2 \times m \times \frac{1}{m} = p^{2} - 2 \times p \times 2 + 2^{2}$

$\implies m^{2} + \frac{1}{m^{2}} - 2= p^{2} - 4p + 4$

$\implies m^{2} + \frac{1}{m^{2}} = p^{2} - 4p + 4+2$

$\implies \green {m^{2} + \frac{1}{m^{2}} = p^{2} - 4p + 6 }$

$Hence\: Proved$

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