Question

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