Is something sketchy going on here?

Factor the second equation:

$ab(2a^2+3b^2-5)=0$

First we have $a=0$ . Substitute that in the first equation to obtain:

$9b^4-30b^2+1=0$

That equation has two positive roots: $b=\dfrac{3 \pm \sqrt{6}}{3}$

Now with $b=0$ do the same:

$4a^4-20a^2+1=0$

That equation has two positive roots: $a=\dfrac{\sqrt{6} \pm 2}{2}$

Finally, do the same with $2a^2+3b^2=5$ . Rearrange the first equation:

$(2a^2+3b^2)^2+24a^2b^2=10(2a^2+3b^2)-1$

$5^2+24a^2b^2=10(5)-1$

$a^2b^2=1 \implies a^2=\dfrac{1}{b^2}$

Substitute that in $2a^2+3b^2=5$ :

$\dfrac{2}{b^2}+3b^2=5 \implies 3b^4-5b^2+2=0$

That equation has two positive roots: $b=1$ and $b=\dfrac{\sqrt{6}}{3}$ . For $a$ we get, respectively: $a=1$ and $a=\dfrac{\sqrt{6}}{2}$ .

Finally, our pairs are: $(a,b)=\left(0,\dfrac{3+\sqrt{6}}{3}\right),\left(0,\dfrac{3-\sqrt{6}}{3}\right),\left(\dfrac{\sqrt{6}+2}{2},0\right),\left(\dfrac{\sqrt{6}-2}{2},0\right),(1,1),\left(\dfrac{\sqrt{6}}{2},\dfrac{\sqrt{6}}{3}\right)$

Sum everything and you get: $\dfrac{24+11\sqrt{6}}{6}$ . So, the final answer is $24+11+6+6=\boxed{47}$ .

Multiply the second equation by $4\sqrt{6}$ , add it to the first, then factor to get $(a\sqrt{2}+b\sqrt{3})^4=10(a\sqrt{2}+b\sqrt{3})^2-1.$ Letting $x=a\sqrt{2}+b\sqrt{3}$ and rearranging gives $x^4-10x^2+1=0$ Which can be solved as a quadratic in $x^2$ after which $x$ can be computed as $\pm\sqrt{2}\pm\sqrt{3}$ , after which the solutions are easy to find.

$ab(2a^2+3b^2)=5ab$ $ab=0 or 2a^2+3b^2=5$ $(2a^2+3b^2)^2+24a^2b^2=10(2a^2+3b^2)-1=50-1=49$ $25+24a^2b^2=49$ $24a^2b^2=24$ $a^2b^2=1$ $2a^2+3b^2=5$ $2a^2+3(\frac{1}{a^2})=5$ $2a^4+3=5a^2$ $a^2=\frac{3}{2}, b^2=\frac{2}{3}$ $ab=0 <=> a=0, b=0$ $a=0, 9b^4=30b^2-1$ $9b^4-30b^2+1=0$ $b^2=\frac{+-\sqrt{216}+15}{9}$ $b_1=\frac{\sqrt{+\sqrt{216}+15}}{3}$ $b_2=\frac{\sqrt{-\sqrt{216}+15}}{3}$ $b_1+b_2=\frac{1}{3}(\sqrt{\sqrt{216}+15}+\sqrt{-\sqrt{216}+15})=$ $\frac{1}{\sqrt{-\sqrt{216}+15}}+\frac{1}{\sqrt{+\sqrt{216}+15}}=$ $\frac{1}{\sqrt{-6\sqrt{6}+15}}+\frac{1}{\sqrt{6\sqrt{6}+15}}$ $b=0$ $4a^4-20a^2+1=0$ $a^2=\frac{+-\sqrt{96}+10}{4}$ $a_1=\frac{\sqrt{+\sqrt{96}+10}}{2}$ $a_2=\frac{\sqrt{-\sqrt{96}+10}}{2}$ $S=2+\frac{5}{\sqrt{6}}+\frac{1}{\sqrt{-6\sqrt{6}+15}}+\frac{1}{\sqrt{+6\sqrt{6}+15}}+\frac{1}{\sqrt{+4\sqrt{6}+10}}+\frac{1}{\sqrt{-4\sqrt{6}+10}}$ $=2+\frac{5}{\sqrt{6}}+\frac{1}{\sqrt{3}*(\sqrt{3}-\sqrt{2})}+\frac{1}{\sqrt{3}*(\sqrt{3}+\sqrt{2})}+\frac{1}{\sqrt{2}*(\sqrt{3}+\sqrt{2})}+\frac{1}{\sqrt{2}*(\sqrt{3}-\sqrt{2})}$ $=\frac{2\sqrt{6}+5+\sqrt{2}(\sqrt{3}+\sqrt{2})+\sqrt{2}(\sqrt{3}-\sqrt{2})+\sqrt{3}(\sqrt{3}-\sqrt{2})+\sqrt{3}(\sqrt{3}+\sqrt{2})}{\sqrt{6}}$ $=\frac{2\sqrt{6}+6+(\sqrt{3}+\sqrt{2})^2}{\sqrt{6}}$ $=\frac{2\sqrt{6}+6+5+2\sqrt{6}}{\sqrt{6}}$ $=\frac{11+4\sqrt{6}}{\sqrt{6}}$ $=\frac{\sqrt{6}(11+4\sqrt{6})}{6}=\frac{24+11\sqrt{6}}{6}$ $a+b+c+d=\boxed{47}$

From the second equation, I found that a =0 or b = 0 or 2.a^2+3.b^2 = 5.

If a = 0, from the first equation, I have 9.b^4-30.b^2+1=0.

If b = 0, from the first equation, I have 4.a^4-20.a^2+1=0.

If 2.a^2+3.b^2=5, I have a=b=1 or a= \frac{3}{2} .b .