If $2a +\dfrac{1}{2a}= 8$, find:

(i) $4a^2 +\dfrac{1}{4a^2}$

(ii) $16a^4 +\dfrac{1}{16a^4}$

(i) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

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

Putting the value $2a + \dfrac{1}{2a} = 8$, we get

$8^2 = 4a^2 + 2 + \dfrac{1}{4a^2}\\[1em] ⇒ 64 = 4a^2 + 2 + \dfrac{1}{4a^2}\\[1em] ⇒ 4a^2 + \dfrac{1}{4a^2} = 64 - 2 \\[1em] ⇒ 4a^2 + \dfrac{1}{4a^2} = 62$

Hence, the value of $4a^2 + \dfrac{1}{4a^2}$ is 62.

(ii) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

$\Big(4a^2 + \dfrac{1}{4a^2}\Big)^2 = (4a^2)^2 + 2 \times 4a^2 \times \dfrac{1}{4a^2} + \Big(\dfrac{1}{4a^2}\Big)^2\\[1em] ⇒ \Big(4a^2 + \dfrac{1}{4a^2}\Big)^2 = 16a^4 + 2 + \dfrac{1}{16a^4}$

Putting the value $4a^2 + \dfrac{1}{4a^2} = 62$, we get

$62^2 = 16a^4 + 2 + \dfrac{1}{16a^4}\\[1em] ⇒ 3,844 = 16a^4 + 2 + \dfrac{1}{16a^4}\\[1em] ⇒ 16a^4 + \dfrac{1}{16a^4} = 3,844 - 2 \\[1em] ⇒ 16a^4 + \dfrac{1}{16a^4} = 3,842$

Hence, the value of $16a^4 + \dfrac{1}{16a^4}$ is 3,842.