Bashing is for the weak, WE factor

What I did was I formed an equation in terms of c.
Let the two roots of the equation be w,z. Then I simply calculated the sum of w and z from the equation which comes out to be -22/45 . since we know that a=3c. For two values of c i.e. w and z, there will be two values of a- u and x. Therefore u+x= 3(w +z)
Similarly b= 6c . Therefore v+y= 6(w+z). Thus in the end the sum turns out to be 10(w+z)= -44/9. Thus p+q= 53

For the equation: a=3c , b= 6c Put the above values of a and b in the first equation.

Sorry for the bad presentation. I don't know how to type in Latex.

This is not a very in-depth solution.

The first equation can be factored into $(a+1)^2=(2b+3)^2$ . Which after square rooting becomes $(a+1)=\pm(2b+3)$ . THE $\pm$ IS VERY IMPORTANT. It is why we have 6 answers and not 3.

The second can be factored into $8(b-2c)^3=(b+2c)^3$ .

The third can be factored into $(a-3c)^4=0$ .

Thus we get $a=(-2b-4,2b+2), b=6c, c=\frac{a}{3}$ . The reason there are 2 solutions for each variable is that there are two a's, which means that both have to be plugged into the equation and two answers will result for each variable. Plugging these terms back into the original equation yields $a=(-\frac{4}{5},-\frac{2}{3}), b=(-\frac{8}{5},-\frac{4}{3}), c=(-\frac{4}{15},-\frac{2}{9})$ .

Some simple arithmetic leads us to $-\frac{p}{q}=-\frac{44}{9}$ . Thus $p+q=53$