How many solutions are for the quadratic equation $(12+a)x^2+12ax+9a=0$

Question

I need to give a result for each $a\in \mathbb{R}$ how many solution is there to this equation:

$(12+a)x^2+12ax+9a=0$.

My attempt: Check for $B^2-4AC:$

$B^2-4AC = 144a^2-4(108a+9a^2)=108a^2-432a=108a(a-4)$

I feel like it leads me nowhere..

Thanks guys in advance :)

Answer 1

You are on the right track .. Now whenever $B^2-4ac < 0$, the quadratic equation has no real roots so you're answer is whenever $108a(a-4) <0$ i.e $0 \leq a <4$

Answer 2

Hint:

The discriminant is a quadratic polynomial in $a$, and you have a theorem on the sign of a quadratic polynomial: it has the sign of its leading coefficient, except between its roots, if any.

• if $\Delta>0$, two real roots;
• if $\Delta=0$, one (real) double root;
• if $\Delta<0$, two complex conjugate roots.