help review question!!!!

Each quadratic can have two real roots, so the maximum number is 6 real roots.

i answered that but it was wrong.... help! Any other answers I can achieve? it wasn't 6 

Each of these quadratics have a maximum of 2 solutions as long as their discriminant works.

try creating a system of inequalities using the properties of a discriminant with a, b, c??????

Not sure if that helped I don't have much time to solve this because I have a math class in literally 5 min.

thanks! the trick actually was that a,b,c were positive. the answer was 4

Mmm

ax^2+bx+c,  bx^2+cx+a,  cx^2+ax+b

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-c \pm \sqrt{c^2-4ab} \over 2b}\\ x = {-a \pm \sqrt{a^2-4cb} \over 2c}\\ \)

These are the zeros for the 3 expressions.  Off hand I cannot see why there cannot be 6 roots in total.......

Ok I graphed it in desmos and if does not appear to be able to have 6 roots.

I can get 4 but I could work out how to get more.

https://www.desmos.com/calculator/hrlzvwageo

For any of these to have two roots, the desciminant must be greater than 0.

That can certainly be true for at least one of these but can it be true of 2 or for all there.

Obviously it can be true for 2 cause I graphed it.

So I really do not have a definitive answer.    Most likely 4 but I have not seen any maths to prove it.