Question

If all values of $p$ so that $6$ lies between roots of the equation $x^2+2(p-3)x+9=0$, belong to the interval $(-\infty ,-\frac{a}{a+1})$, then find $a$.

Answer 1

Let $f\left(x\right)=x^2+2(p-3)x+9$.
As 6 lies between the roots of $f\left(x\right)=0$, we have $D>0$ and $af\left(6\right)<0$
(I) Consider $D>0$ $(2(p-3){)}^2-\mathrm{4.1.9}>0$
$\Rightarrow {(p-3)}^2-9>0\Rightarrow p(p-6)>0$
$\Rightarrow p\in (-\infty ,0)\cup (6,\infty )$
(II) Consider $af\left(6\right)<0$ $(36+12(p-3)+9)<0$
$\Rightarrow 12p+9<0\Rightarrow p+\frac{3}{4}<0$ $\Rightarrow p\in (-\infty ,-\frac{3}{4})$
Hence, the value of $p$ satisfying both inequalities at the same time are $p\in (-\infty ,-\frac{3}{4})$.