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# Help asap

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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 $a + b + c + d$?

Jan 24, 2018

### 1+0 Answers

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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   Jan 24, 2018