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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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Numerical Ability
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gatecse
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Dec 7, 2017
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pavan singh

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Best answer
$\quad \left(Z+\dfrac{1}{Z}\right)^{2}$
$= \left(z^{2} + 2(z)\left(\dfrac{1}{z}\right) + \left(\dfrac{1}{z}\right)^{2}\right) = \left(z^{2} + \dfrac{1}{z^{2}}\right)+2 =98$
$\Rightarrow 982 = 96$ is answer..
answered
Dec 15, 2014
by
Card Wizard
edited
Dec 7, 2017
by
pavan singh
comment
+1
answer is 96
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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 5, 2019
by
bhupendrakumar
edited
Sep 5, 2019
by
Lakshman Patel RJIT
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Answer:
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