Question

The probability of selecting integers$a\in [-5,30]$, such that $x^2+2(a+4)x-5a+64>0$ for all $x\in R$is:

Answer 1

Solve the inequation for finding the number of integers:

$x^2+2(a+4)x–(5a–64)>0$

Let $y=x^2+2(a+4)x-(5a-64)$

$\Rightarrow y>0$

This is possible only if the discriminant $D<0$

Therefore $4{(a+4)}^2+4(5a–64)<0$

$\Rightarrow a^2+13a–48<0$

$\Rightarrow (a+16)(a–3)<0$

So, $a\in (-16,3)$

Total integers are $18$ here

But we are given that $a\in [-5,30]$ so , total out comes are $36$

And favorable out comes will be in $(-16,3)\cap [-5,30]=[-5,3)$

So favorable out comes are $8$

Therefore Probability $=\frac{8}{36}=\frac{2}{9}$

Hence, the required probability is $\frac{2}{9}$