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.

  • $\begingroup$ Asking at Stack Exchange is a computer method ;-), and it's not always better than asking at Wolfram Alpha. $\endgroup$ – Professor Vector Sep 11 '17 at 12:46
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    $\begingroup$ The quartic factors. $\endgroup$ – i. m. soloveichik Sep 11 '17 at 12:55

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

$$25 - 5(u+1)^2 - 2(u+1)^4 = 0$$

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, $x = t^2 = \frac{14+4\sqrt{10}}{9}.$

  • $\begingroup$ Thanks. I have now understood the method of attack. $\endgroup$ – rugi Sep 11 '17 at 15:12
  • $\begingroup$ @GAVD How did you get $25-5(u+1)^2-2(u+1)^4=0$? Explain please. Thank you! $\endgroup$ – Michael Rozenberg Sep 11 '17 at 15:15
  • $\begingroup$ Well, this is one of the way of attack the 4-degrees equations. In fact, I go from the form $k(at^2+bt+c)^2 + l(at^2+bt+c) + m = 0$, but it is really ugly, but nice for $\frac{1}{t}$. $\endgroup$ – GAVD Sep 11 '17 at 15:27

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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    $\begingroup$ @rugi My solution it's a solution by "pencil and paper", without computer of course.. $\endgroup$ – Michael Rozenberg Sep 11 '17 at 15:31
  • $\begingroup$ can you prove your Statement Michael? $\endgroup$ – Dr. Sonnhard Graubner Sep 11 '17 at 19:42
  • $\begingroup$ About which statement do you say? $\endgroup$ – Michael Rozenberg Sep 11 '17 at 19:45
  • $\begingroup$ this here Michael: "My solution it's a solution by "pencil and paper", without computer of Course" $\endgroup$ – Dr. Sonnhard Graubner Sep 11 '17 at 19:49
  • $\begingroup$ Sonnhard Just see my solution and try to restore it by yourself. $\endgroup$ – Michael Rozenberg Sep 11 '17 at 19:50

try the Ansatz $$18t^4-18t^3-17t^2-8t-2=(at^2+bt+c)(dt^2+et+f)$$


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