Question

If the roots of the equation $x^2-8x+a^2-6a=0$ are real and distinct then find number of all possible integral values of $a$

Answer 1

Since the roots of the given equation are real and distinct, we must have $D>0$
$\Rightarrow 64-4(a^2-6a)>0\Rightarrow 4[16-a^2+6a]>0$
$-4(a^2-6a-16)>0$

$\Rightarrow a^2-6a-16<0$
$\Rightarrow (a-8)(a+2)<0$

$\Rightarrow -2<a<8$
Hence, the roots of the given equation are real if '$a$' lies between $-2$ and $8$.