Question

If the roots of the equation $x^2-8x+m(m-6)=0$ are real distinct, then find all possible values of $m$

Answer 1

Consider $x^2-8x+m(m-6)=0$

For the roots to be real and distinct, the discriminant must be greater than or equal to 0.

That is $b^2-4ac\ge 0$

$\Rightarrow (-8{)}^2-4(1\left)\right[m(m-6)]\ge 0$

$\Rightarrow 64-4m(m-6)\ge 0$

$\Rightarrow 64\ge 4m(m-6)$

$\Rightarrow m(m-6)\le 16$

$\Rightarrow m^2-6m-16\le 0$

$\Rightarrow (m-8)(m+2)\le 0$

$\Rightarrow -2\le m\le 8$