### Instruction: Find the real roots of
\[\frac{6 - x}{x^2 - 4} = 2 + \frac{x}{x + 2}.\]Enter the roots, separated by commas.### Response: To find the real roots of this equation, I want to manipulate it into a form where I can use the quadratic formula or factorization.
I notice that the denominator of the left side, \(x^2 - 4\), can be factored as \((x + 2)(x - 2)\).
This means that the denominator of the right side, \(x + 2\), is a factor of the denominator of the left side.
This suggests that I can simplify the equation by multiplying both sides by \(x + 2\).
Doing so, I get
\[\frac{6 - x}{x - 2} = 2(x + 2) + x.\]
Now, I want to clear the fraction by multiplying both sides by \(x - 2\).
This gives me
\[6 - x = 2(x + 2)(x - 2) + x(x - 2).\]
Expanding the parentheses, I get
\[6 - x = 2x^2 - 8 + x^2 - 2x.\]
Collecting like terms, I get
\[-3x^2 + x + 14 = 0.\]
This is a quadratic equation that I can solve by using the quadratic formula.
The quadratic formula tells me that the roots are
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
where \(a = -3, b = 1, c = 14\).
Plugging in these values, I get
\[x = \frac{-1 \pm \sqrt{1 - 4(-3)(14)}}{-6}.\]
Simplifying the radical, I get
\[x = \frac{-1 \pm \sqrt{169}}{-6}.\]
The square root of 169 is 13, so I get
\[x = \frac{-1 \pm 13}{-6}.\]
This gives me two possible values for x:
\[x = \frac{-1 + 13}{-6} = -2\]
and
\[x = \frac{-1 - 13}{-6} = \frac{7}{3}.\]
These are the real roots of the equation.