Question

If the roots of the equation $x^2-8x+a^2-6a=0$ are real and distinct, then the number of integral value(s) of $a$ is

Answer 1

$x^2-8x+a^2-6a=0$
Since, roots are real and distinct.
$\therefore D>0\Rightarrow 64-4(a^2-6a)>0\Rightarrow 16-a^2+6a>0\Rightarrow a^2-6a-16<0\Rightarrow (a-8)(a+2)<0\Rightarrow a\in (-2,8)\Rightarrow a=-1,0,1,2,3,4,5,6,7$

Hence, the number of integral values of $a$ is $9.$