(a) Suppose $m$ is an integer so that $(m^2 - 1)x^2 - 3(3m - 1)x + 18 = 0$ has two positive integer roots. Find $m.$
(b) Now, suppose that we have a triangle $ABC$ with sides $a,b,c$ such that \begin{align*} c &= 2 \sqrt{3} \\ m^2 + a^2m - 8a &= 0 \\ m^2 + b^2m - 8b &= 0. \end{align*} Find the area of $ABC.$
For part (a), I first attempted to apply the discriminant to find $m,$ which gave me $m = 2.$ However, for part (b), I want to apply Heron's Formula, but I don't know how to find $a$ and $b.$ Can someone give me a hint please?
a) $$(m^2 - 1)x^2 - 3(3m - 1)x + 18=((m-1)x-3)((m+1)x-6),$$ which since $m^2-1\neq0$, gives $$x=\frac{3}{m-1}$$ or $$x=\frac{6}{m+1}$$ and easy to see that only $m=2$ is valid.
b) Since $a^2-4a+2=0$ and $b^2-4b+2=0$, we obtain following triples for sides-lengths of the triangle: $$(2+\sqrt2,2+\sqrt2,2\sqrt3)$$ and $$(2+\sqrt2,2-\sqrt2,2\sqrt3).$$ Now, for calculating of the area we can use the following formula: $$S_{\Delta ABC}=\frac{1}{4}\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4}.$$ For the first we obtain: $$S=\sqrt{9+12\sqrt2}.$$ For the second we obtain $S=1.$