If a² - 3a + 1 = 0 and a ≠ 0; find :

(i) $a + \dfrac{1}{a}$

(ii) $a^2 + \dfrac{1}{a^2}$

(i) Given,

⇒ a² - 3a + 1 = 0

⇒ a² + 1 = 3a

Dividing above equation by a, we get :

$\Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{3a}{a}$

$\Rightarrow a + \dfrac{1}{a} = 3$

Hence, $a + \dfrac{1}{a} = 3$.

(ii) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow 3^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow a^2 + \dfrac{1}{a^2} = 9 - 2 \\[1em] \Rightarrow a^2 + \dfrac{1}{a^2} = 7.$

Hence, $a^2 + \dfrac{1}{a^2} = 7$.