How do you solve $\frac{x}{x + 4} = \frac{2}{x}$ $x/(x+4)=2/x$ and check for extraneous solutions?

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Answer 1 · 2017-04-09T11:22:54

$x \in {4, - 2}$ $x in {4, -2}$

Multiply both sides.

$\frac{x}{x + 4} \cdot x (x + 4) = \frac{2}{x} \cdot x (x + 4)$ $x/(x+4) * x(x+4) = 2/x * x(x+4)$

$x^{2} = 2 x + 8$ $x^2 = 2x + 8$

$x^{2} - 2 x - 8 = 0$ $x^2 - 2x - 8 = 0$

Sum $= 2$ $= 2$ , Product = $- 8$ $-8$ , therefore $x = 4 or x = - 2$ $x = 4 or x = -2$

The extraneous ones would be $x = 0 or x = - 4$ $x = 0 or x = -4$ , because there would be zero denominator.

How do you solve xx+4=2xx/(x+4)=2/x and check for extraneous solutions?