If $x -\dfrac{1}{x}= 3$, the value of $x^2 +\dfrac{1}{x^2}$ is:

1. 0

2. 7

3. 11

4. 9

Using the formula,

[∵(x - y)² = x² - 2xy + y²]

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

Putting the value, $x -\dfrac{1}{x}= 3$

$(3)^2 = x^2 - 2 + \dfrac{1}{x^2}\\[1em] ⇒ 9 = x^2 - 2 + \dfrac{1}{x^2}\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 9 + 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 11$

Hence, option 3 is the correct option.