If $a^2 +\dfrac{1}{a^2}= 23$, find: $a +\dfrac{1}{a}$

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 $a^2 +\dfrac{1}{a^2} = 23$,we get

$⇒ \Big(a + \dfrac{1}{a}\Big)^2 = \Big(a^2 + \dfrac{1}{a^2}\Big) + 2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = 23 + 2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = 25\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big) = \sqrt{25}\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big) = 5 \text{ or} -5$

Hence, the values of $\Big(a + \dfrac{1}{a}\Big)$ are 5 or -5.