GATE CS 2014 Set 1 QGA-4 — General Aptitude

We are given the equation:

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

Using the algebraic identity $(a + b)^2 = a^2 + b^2 + 2ab$, we can expand the left-hand side by setting $a = z$ and $b = \frac{1}{z}$:

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

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

Substitute the given value $\left(z + \frac{1}{z}\right)^{2} = 98$ into the expanded equation:

$$98 = z^2 + \frac{1}{z^2} + 2$$

Solving for $\left(z^2 + \frac{1}{z^2}\right)$:

$$z^2 + \frac{1}{z^2} = 98 - 2 = 96$$