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Problem: What is the sum of the solutions to the equation: $\sqrt[4]{x} = \frac{12}{7-\sqrt[4]{x}}$

Attempt:

$(\sqrt[4]{x})({7-\sqrt[4]{x}}) = 12$

$7(\sqrt[4]{x}) - (\sqrt[4]{x})^2 = 12$

$[7x^{\frac{1}{4}} - x^{\frac{1}{2}} = 12]1^4$

$2401x - x^2 - 20736 = 0$.

Where in the roots are 2392.33 and 8.667. I stopped there as I know what I'm doing is wrong. By using a calculator, solving for x results to 256 and 81 when added equals to 337 which is the answer. What part of manual solving did I get wrong? Thank you~

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    $\begingroup$ Ooo, careful. When you raise the left hand side to the fourth power, you get some extra terms, for the same reason that $(x+y)^2 \ = \ x^2 + 2xy + y^2 \ \neq \ x^2 + y^2$. $\endgroup$ – Kaj Hansen Jun 20 '18 at 8:37
  • $\begingroup$ Thank you, didn't notice that. Looks like I don't improve in math at all. :( $\endgroup$ – Jayce Jun 20 '18 at 9:48
  • $\begingroup$ We all make mistakes. If only you could see how many deleted posts I have on this site ;) $\endgroup$ – Kaj Hansen Jun 20 '18 at 9:53
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HINT

Let $\sqrt[4]{x}=t$ then

$$t=\frac{12}{7-t}\iff t^2-7t+12=0$$

and solve for $t$, then for the solutions $t_0>0$ solve $\sqrt[4]{x}=t_0$.

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  • $\begingroup$ Didn't know we could do that. The roots of $t^2-7t+12=0$ are 4 and 3 and when substituted to $\sqrt[4]{x}=t$ will result to 256 and 81. When added equals to 337. But any clue why my solution didn't work? I think solving like that is still plausible. $\endgroup$ – Jayce Jun 20 '18 at 8:40
  • $\begingroup$ @Jayce As already pointed out the mistake is in the last step. $\endgroup$ – user Jun 20 '18 at 9:12
  • $\begingroup$ Thank you, I appreciate it. $\endgroup$ – Jayce Jun 20 '18 at 9:27
  • $\begingroup$ @Jayce You are welcome! Bye $\endgroup$ – user Jun 20 '18 at 9:30

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