JEE Main 2021 — Quadratic Equation Question with Solution

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,

$D<0$

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

$\Rightarrow a^2+16+8a+5a-64<0$

$\Rightarrow a^2+13a-48<0$

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

$\Rightarrow a\in \left(-16,3\right)$

$\therefore$ Possible values for $a$ are $\left\{-15, -14, \dots \dots .,2\right\}$ containing $18$ integers.

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.

And, $\left[-5, 30\right]$ has $36$ numbers.

$\therefore$ Required probability

$=\frac{8}{36}$

$=\frac{2}{9}$