What is the sum of the solutions to the following equation? \(\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}\).

Mathgenius Dec 9, 2018

#1**+1 **

It should be \(81+256=337.\) First, change the fourth root into \(\frac{1}{4}.\)

Now, without words:

\(x^{\frac{1}{4}}=\frac{12}{7-x^{\frac{1}{4}}}\)

\(x^{\frac{1}{4}}\left(7-x^{\frac{1}{4}}\right)=\frac{12}{7-x^{\frac{1}{4}}}\left(7-x^{\frac{1}{4}}\right)\)

\(x^{\frac{1}{4}}\left(7-x^{\frac{1}{4}}\right)=12\)

Expand, \(7x^{\frac{1}{4}}-\sqrt{x}=12\)

\(7x^{\frac{1}{4}}-\left(x^{\frac{1}{4}}\right)^2=12\)

Plug variables instead of \(x^\frac{1}{4}\)

to get \(x=81,\:x=256\)

Thus, te answer is \(81+256=\boxed{337}.\)

.tertre Dec 9, 2018

#2**+2 **

Let's see hoe tertre arrived at his solution

7x^(1/4) - ( x^1/4)^2 = 12 multiply through by -1

(x^1/4)^2 - 7(x^1/4) = -12

(x^1/4)^2 - 7(x^1/4) + 12 = 0 let x^1/4 = a and we have

a^2 - 7a + 12 = 0 factor

(a - 4) (a - 3) = 0

Setting both factors to 0 and solving for "a" gives a =4 or a = 3

So

x^1/4 = 4 or x^1/4 = 3 take each side to the 4th power

x = 256 or x = 81

And the sum of these is 337.....just as tertre found !!!!!

CPhill Dec 9, 2018