[Math] the sum of the solutions to the equation

algebra-precalculus

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~

Best Answer

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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