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**HallsofIvy** Equivalently:

$\displaystyle \frac{x^2- 4x}{x^2+ 2}= p$

$\displaystyle x^2- 4x= px^2+ 2p$

$\displaystyle (1- p)x^2- 4x= 2p$

$\displaystyle x^2- \frac{4}{1- p}x= \frac{2p}{1- p}$

Complete the square:

$\displaystyle x^2- \frac{4}{1- p)x+ \frac{4}{(1- p)^2= \frac{2p}{1- p}+ \frac{4}{(1- p)^2= \frac{2p- 2p^2+ 4}{(1- p)^2}$

$\displaystyle (x- \frac{2}{1- p})^2= -2\frac{p^2- p- 2}{(1- p)^2}$

Solve by taking the square root of both sides. If the right side is negative, there is no root. If the right side is positive we can take both "+" and "-" and there are two real roots. If the right side is 0, there is only one root.