Pre-Calculus Review Problems | Solutions 1 Algebra and Geometry

(d) Note that any solution xcannot be equal to 2 or 2. x 1 x 2 + 2x+ 1 x+ 2 = 0 x 1 x 2 x+ 2 x+ 2 + 2x+ 1 x+ 2 x 2 x 2 = 0 x2 + x 2 x2 4 2x2 3x 2 x2 4 = 0 3x2 2x 4 x2 4 = 0: The only way for this equation to be true is if the numerator on the left-hand side is 0, which occurs


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