<s>[INST] Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$ 
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$ 

Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing seem to work. Then putting the original equation in wolfram alpha I got solution $x=\frac{2}{9}(7+2 \sqrt{10})$. Can anyone suggest a method to solve it without wolfram alpha or any such computer method. Thanks in advance.
 [/INST] Let $\sqrt{x}=\frac{t}{6}$.
Thus, we need to solve
$$18\cdot\frac{t^4}{1296}-18\cdot\frac{t^3}{216}-17\cdot\frac{t^2}{36}-8\cdot\frac{t}{6}-2=0$$ or
$$t^4-6t^3-34t^2-96t-144=0$$ or for all real $k$
$$(t^2-3t+k)^2-9t^2-k^2-2kt^2+6kt-34t^2-96t-144=0$$ or
$$(t^2-3t+k)^2-((2k+43)t^2-(6k-96)t+k^2+144)=0.$$
Now, we'll choose a value of $k$ such that
$$(2k+43)t^2-(6k-96)t+k^2+144=(at+b)^2,$$
for which we need
$$9(k-16)^2-(2k+43)(k^2+144)=0.$$
Easy to see that $k=-9$ is valid.
Id est, we need to solve
$$(t^2-3t-9)^2-(25t^2+150t+225)=0$$ or
$$(t^2-3t-9)^2-25(t+3)^2=0$$ or
$$(t^2-3t-9-5(t+3))(t^2-6t-9+5(t+3))=0$$ or
$$(t^2-8t-24)(t^2+2t+6)=0$$ or
$$t^2-8t-24=0$$ or
$$(t-4)^2=40,$$ which gives
$$6\sqrt{x}=4+\sqrt{40}$$ or
$$\sqrt{x}=\frac{2+\sqrt{10}}{3}$$ or
$$x=\frac{14+4\sqrt{10}}{9}.$$
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