12 votes 12 votes If $\large\left(z + \dfrac{1}{z}\right)^{2}= 98$, compute $\large \left(z^{2} +\dfrac{1}{z^{2}}\right)$. Quantitative Aptitude gatecse-2014-set1 quantitative-aptitude easy numerical-answers numerical-computation + – gatecse asked Sep 15, 2014 edited Dec 6, 2017 by pavan singh gatecse 3.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 24 votes 24 votes $\quad \left(Z+\dfrac{1}{Z}\right)^{2}$ $= \left(z^{2} + 2(z)\left(\dfrac{1}{z}\right) + \left(\dfrac{1}{z}\right)^{2}\right) = \left(z^{2} + \dfrac{1}{z^{2}}\right)+2 =98$ $\Rightarrow 98-2 = 96$ is answer.. Card Wizard answered Dec 15, 2014 edited Dec 6, 2017 by pavan singh Card Wizard comment Share Follow See 1 comment See all 1 1 comment reply Regina Phalange commented Apr 6, 2017 i moved by Lakshman Bhaiya Sep 5, 2019 reply Follow Share answer is 96 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes $\left(Z+\dfrac{1}{Z}\right)^{2}=98$ $\implies Z^{2}+\dfrac{1}{Z^{2}}+2Z\cdot \dfrac{1}{Z}=98$ $\implies Z^{2}+\dfrac{1}{Z^{2}}+2=98$ $\implies Z^{2}+\dfrac{1}{Z^{2}}=96$ bhupendrakumar answered Sep 4, 2019 edited Sep 5, 2019 by Lakshman Bhaiya bhupendrakumar comment Share Follow See all 0 reply Please log in or register to add a comment.