# What is the sum of the solutions to the equation

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~

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

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 $\sqrt[4]{x}=t_0$.

• 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. – Jayce Jun 20 '18 at 8:40
• @Jayce As already pointed out the mistake is in the last step. – user Jun 20 '18 at 9:12
• Thank you, I appreciate it. – Jayce Jun 20 '18 at 9:27
• @Jayce You are welcome! Bye – user Jun 20 '18 at 9:30