The probability of selecting integers a ∈ [- 5, 30] such that x ^ 2...

We have,

${x}^{2} + 2\left(a + 4\right)x - 5a + 64 > 0$

If $A > 0 and D = {B}^{2} - 4AC < 0$, then $A{x}^{2} + Bx + C > 0 \forall x \in R$.

Hence,

$D < 0$

$⇒ 4{\left(a + 4\right)}^{2} - 4\left(- 5a + 64\right) < 0$

$⇒ {a}^{2} + 16 + 8a + 5a - 64 < 0$

$⇒ {a}^{2} + 13a - 48 < 0$

$⇒ \left(a + 16\right)\left(a - 3\right) < 0$

$⇒ a \in \left(- 16, 3\right)$

$∴$ Possible values for $a$ are \left\{- 15, - 14, \ldots \ldots . , 2\right\}$containing$18$integers.</p><p>But given range of$a$is \left[- 5, 30\right]$, hence $a$ would take values \left\{- 5, - 4, - 3, - 2, - 1, 0, 1, 2\right\}$, containing$8$integers.</p><p>And, \left[- 5, 30\right]$ has $36$ numbers.

$∴$ Required probability

$= \dfrac{8}{36}$

$= \dfrac{2}{9}$