If a is a positive and $a^2 +\dfrac{1}{a^2}= 18$; then the value of $a -\dfrac{1}{a}$ is:

1. 4

2. 16

3. 20

4. 2√5

Using the formula,

[∵(x - y)² = x² - 2xy + y²]

$\Big(a -\dfrac{1}{a}\Big)^2 = a^2 - 2 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\\[1em] = a^2 - \dfrac{2a}{a} + \Big(\dfrac{1}{a}\Big)^2\\[1em] = a^2 - 2 + \Big(\dfrac{1}{a}\Big)^2\\[1em]$

Putting the value, $a^2 +\dfrac{1}{a^2}= 18$

$⇒ \Big(a -\dfrac{1}{a}\Big)^2 = a^2 + \Big(\dfrac{1}{a}\Big)^2 - 2\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big)^2 = 18 - 2\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big)^2 = 16\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big) = \sqrt{16}\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big) = 4 \text{ or} -4$

As a is a positive number.

So, $\Big(a -\dfrac{1}{a}\Big) = 4$

Hence, option 1 is the correct option.