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There are two pairs $(x,y)$ of real numbers that satisfy the equation $x+y = 3xy = 4$. Given that the solutions $x$ are in the form $x = \frac{a \pm b\sqrt{c}}{d}$ where $a$, $b$, $c$, and $d$ are positive integers and the expression is completely simplified, what is the value of $a + b + c + d$?

Guest Jan 24, 2018
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\( $x = \frac{a \pm b\sqrt{c}}{d}$\)

 

x + y   = 4       (1)

3xy  = 4  ⇒   y  = 4/ [ 3x]     (2)

 

Sub (2)  into (1)

 

x + 4 / [3x]  =   4            multiply through by  3x

 

3x^2  + 4  =  12x        rearrange as

 

3x^2  - 12x  + 4  =  0

 

x =  [ 12 ±√ [ 144 - 48] ] / 6

 

x  =   [ 12 ±√ 96 ] / 6 

 

x = [  12 ± 4√6 ] / 6      =   [ 6 ± 2√6 ] / 3

 

So  

 

A =  6     B  = 2   C   = 6   D   =  3

 

And their sum is 

 

17 

 

 

cool cool cool

CPhill  Jan 24, 2018

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