Question

How could this Rational Equation be solved? The answer is supposed to be 4 times the negative or positive square root of 30, and I've tried it a bunch of times, but keep getting a similar, but slightly off answer. Maybe I'm using the wrong LCD?

3/x^2-4 + 1/2 = 4/x-2

Answer 1

$\frac{3}{x^2-4} + \frac{1}{2}= \frac{4}{x-2}\\\\\frac{3}{(x-2)(x+2)} + \frac{1}{2} = \frac{4}{x-2}\\\\\frac{3}{(x-2)(x+2)}= \frac{4}{x-2} - \frac{1}{2}\\\\\frac{3}{(x-2)(x+2)} = \frac{4(x+2)}{(x-2)(x+2)}-\frac{1(x-2)(x+2)}{2(x-2)(x+2)}\\\\\frac{3}{(x-2)(x+2)} = \frac{4x+8}{(x-2)(x+2)}-\frac{x^2-4}{2(x-2)(x+2)}\\\\\frac{3}{(x-2)(x+2)} = \frac{4x+8-x^2+4}{(x-2)(x+2)}\\\\(x-2)(x+2)\times\frac{3}{(x-2)(x+2)} = \frac{4x+8-x^2+4}{(x-2)(x+2)}\times (x-2)(x+2)\\\\3 = 4x + 8 -x^2+4$

$3 = 4x + 8 -x^2+4\\\\3 = 4x+ 12 - x^2\\\\3 = -x^2 + 4x + 12\\\\-x^2 + 4x + 9 = 0\\\\x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\x_{1,2} = \frac{ -(-4) \pm \sqrt{ (-4)^2 - 4 \cdot 1 \cdot (-9)} }{ 2 \cdot 1 }\\\\x_{1,2} = \frac{ 4 \pm \sqrt{ 52 } }{ 2 }\\\\x_1 = \frac{ 4~+~\sqrt{ 52 } }{ 2 } = 2+\sqrt{13}\\\\x_2 = \frac{ 4~-~\sqrt{ 52 } }{ 2 } = 2-\sqrt{13}$

So obviously the answer they gave you is wrong. :P