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If $\large\left(z + \dfrac{1}{z}\right)^{2}= 98$, compute $\large \left(z^{2} +\dfrac{1}{z^{2}}\right)$.
in Numerical Ability by Boss (16.8k points)
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+20 votes
Best answer
$\quad \left(Z+\dfrac{1}{Z}\right)^{2}$

$= \left(z^{2} + 2(z)\left(\dfrac{1}{z}\right) + \left(\dfrac{1}{z}\right)^{2}\right) = \left(z^{2} + \dfrac{1}{z^{2}}\right)+2 =98$

$\Rightarrow 98-2 = 96$ is answer..
by Active (1k points)
edited by
answer is 96
+1 vote

$\implies Z^{2}+\dfrac{1}{Z^{2}}+2Z\cdot \dfrac{1}{Z}=98$

$\implies Z^{2}+\dfrac{1}{Z^{2}}+2=98$

$\implies Z^{2}+\dfrac{1}{Z^{2}}=96$
by (47 points)
edited by
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