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GATE20141GA4
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If $\large\left(z + \dfrac{1}{z}\right)^{2}= 98$, compute $\large \left(z^{2} +\dfrac{1}{z^{2}}\right)$.
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Sep 15, 2014
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$\quad \left(Z+\dfrac{1}{Z}\right)^{2}$
$= \left(z^{2} + 2(z)\left(\dfrac{1}{z}\right) + \left(\dfrac{1}{z}\right)^{2}\right) = \left(z^{2} + \dfrac{1}{z^{2}}\right)+2 =98$
$\Rightarrow 982 = 96$ is answer..
answered
Dec 15, 2014
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Card Wizard
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edited
Dec 6, 2017
by
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$\left(Z+\dfrac{1}{Z}\right)^{2}=98$
$\implies Z^{2}+\dfrac{1}{Z^{2}}+2Z\cdot \dfrac{1}{Z}=98$
$\implies Z^{2}+\dfrac{1}{Z^{2}}+2=98$
$\implies Z^{2}+\dfrac{1}{Z^{2}}=96$
answered
Sep 4
by
bhupendrakumar
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47
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edited
Sep 5
by
Lakshman Patel RJIT
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