Question

Find all possible values of a for which exactly one root of $x^2-(a+1)x+2a=0$ lies in interval $(0,3)\left(0\right)\left(3\right)<0$
$\Rightarrow 2a(9-3(a+1)+2a)<0$
$\Rightarrow 2a(-a+6)<0$
$\Rightarrow a(a-6)>0$
$\Rightarrow a<0ora>6$
Checking the extremes.

Answer 1

Let $f\left(x\right)=x^2-(a+1)x+2a=0$

Let $p$ and $q$ be the roots of the above equation and it is given that it lies between $(0,3)$

$\therefore p=0,q=3$

Given that there exists exactly one root

$\Rightarrow f\left(p\right)f\left(q\right)<0$

$\Rightarrow f\left(0\right)f\left(3\right)<0$

$\Rightarrow [0-(a+1)0+2a][3^2-(a+1)3+2a]<0$

$\Rightarrow 2a[9-3a-3+2a]<0$

$\Rightarrow 2a(6-a)<0$

$\Rightarrow a>0,6-a<0$

$\Rightarrow a>0,-a<-6$

$\Rightarrow a>0,a>6$

$\therefore a\in (-\infty ,0)\cup (6,\infty )$