$\displaystyle \sqrt{\frac{9a+4}{a-6}}$ is a rational number $ \displaystyle \rightleftarrows $ $\displaystyle ( 9a+4)( a-6)$ is a perfect square $\displaystyle ( 9a+4)( a-6) =s^{2} \ \ ( s >0,s\in \mathbb{Z})$ $\displaystyle \rightleftarrows \ ( 9a+3s-25)( 9a-3s-25) =29\cdot 29$ $\displaystyle \rightleftarrows \ 9a+3s-25=-1\ \land \ 9a-3s-25=-841$ $\displaystyle \rightleftarrows \ ( a,s) \ =\ ( -44,\ 140)$ Conclusion: $\displaystyle a=-44$ is the only solution.

Thanks @kaede for the solution,I had let the the denominator and numerator be separate perfect squares but that had no solutions.Loved how you tackled it.

@kaede What does the \land mean in the third to last line? the "^" character? (sorry if the questions kinda dumb)

Alternative solution: We must have $9a+4=x^2, a-6=y^2$, so $x^2-9y^2=58$, thus $(x-3y)(x+3y)=2*29$, and the rest is easy.

TuZo wrote: Alternative solution: We must have $9a+4=x^2, a-6=y^2$, so $x^2-9y^2=58$, thus $(x-3y)(x+3y)=2*29$, and the rest is easy. Or $9a+4=2x^2,a-6=2y^2$ Or $9a+4=29x^2,a-6=29y^2$ Or $9a+4=58x^2,a-6=58y^2$ Or $9a+4=-x^2,a-6=-y^2$ etc

Ok, so moore general: $9a+4=k{{x}^{2}},a-6=k{{y}^{2}}\Rightarrow k(x-3y)(x+3y)=58$, and the rest is easy.

SunnySeattle wrote: @kaede What does the \land mean in the third to last line? the "^" character? (sorry if the questions kinda dumb) It means "and"