The value (values) of x satisfying the equation x² - 6x - 16 = 0 is/are :

1. 8 or -2

2. -8 or 2

3. 8 and -2

4. -8 or 2

Comparing equation x² - 6x - 16 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -6 and c = -16.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times -16}}{2 \times 1} \\[1em] = \dfrac{6 \pm \sqrt{36 + 64}}{2} \\[1em] = \dfrac{6 \pm \sqrt{100}}{2} \\[1em] = \dfrac{6 \pm 10}{2} \\[1em] = \dfrac{6 + 10}{2} \text{ or } \dfrac{6 - 10}{2} \\[1em] = \dfrac{16}{2} \text{ or } \dfrac{-4}{2} \\[1em] = 8 \text{ or } -2.$

Substituting x = 8 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ 8² - 6(8) - 16

⇒ 64 - 48 - 16

⇒ 0.

Substituting x = -2 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ (-2)² - 6(-2) - 16

⇒ 4 + 12 - 16

⇒ 0.

Since, L.H.S. = R.H.S., thus x = -2 is a solution of the equation.

Thus 8 and -2 are the solution of the equation x² - 6x - 16 = 0.

Hence, Option 3 is the correct option.