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avatar+146 

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

 Dec 9, 2018
 #1
avatar+3729 
+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 \(81+256=\boxed{337}.\)

.
 Dec 9, 2018
 #2
avatar+94501 
+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 !!!!!

 

 

cool cool cool

 Dec 9, 2018

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