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.
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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 $$x=\frac{14+4\sqrt{10}}{9}.$$
First, the condition is $x\geq 0$.
Now, as you did, put $t = \sqrt{x}$, then we have the equation. $$18t^4-18t^3 -17t^2-8t-2 = 0.$$
Obviously, $t=0$ is not the solution of this equation. Then, we can divide two sides by $t^4$, $$18-\frac{18}{t}-\frac{17}{t^2}-\frac{8}{t^3}-\frac{2}{t^4} = 0$$ Let $u = \frac{1}{t}$, we have
$$18-18u-17u^2-8u^3-2u^4=0$$
$$25 - 5(u+1)^2 - 2(u+1)^4 = 0$$
Now, you can solve the quadratic equation of $(u+1)^2$:
$$(u+1)^2 = \frac{5}{2}$$ Thus, $u = \frac{\sqrt{5}}{\sqrt{2}}-1 = \frac{\sqrt{10}-2}{2}$.
So, $t = \frac{1}{u} = \frac{\sqrt{10}+2}{3}$. Finally, $x = t^2 = \frac{14+4\sqrt{10}}{9}.$