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 $a + b + c + d$?
+128408 \( $x = \frac{a \pm b\sqrt{c}}{d}$\)
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
