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

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

(ii) $a^4 +\dfrac{1}{a^4}$

(i) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

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

Putting the value $a + \dfrac{1}{a} = 2$,we get

$2^2 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ 4 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 4 - 2 \\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 2$

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

(ii) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

$\Big(a^2 + \dfrac{1}{a^2}\Big)^2 = (a^2)^2 + 2 \times a^2 \times \dfrac{1}{a^2} + \Big(\dfrac{1}{a^2}\Big)^2\\[1em] ⇒ \Big(a^2 + \dfrac{1}{a^2}\Big)^2 = a^4 + 2 + \dfrac{1}{a^4}$

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

$2^2 = a^4 + 2 + \dfrac{1}{a^4}\\[1em] ⇒ 4 = a^4 + 2 + \dfrac{1}{a^4}\\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 4 - 2 \\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 2$

Hence, the value of $a^4 + \dfrac{1}{a^4}$ = 2.