If $a +\dfrac{1}{a}= 3$, find: $a^2 +\dfrac{1}{a^2}$

Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

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

Putting the value $a + \dfrac{1}{a} = 3$,we get

$3^2 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ 9 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 9 - 2 \\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 7$

Hence, the value of $a^2 + \dfrac{1}{a^2}$ is 7.