Question

$(\frac{x-2}{x+2}) ^{2} - 4(\frac{x-2}{x+2}) + 3 = 0$
Solve for x.

An easy question for free points and math practice.
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Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$( { \frac{x - 2}{x + 2} )}^{2} - 4( \frac{x - 2}{x + 2} ) + 3 = 0 \\ = > \frac{{(x - 2)}^{2} }{( {x + 2)}^{2} } - \frac{4(x - 2)}{(x + 2)} + 3 = 0 \\ = > \frac{( {x - 2)}^{2} - 4(x - 2) + 3( {x + 2)}^{2} }{( {x + 2)}^{2} } = 0 \\ = > {x}^{2} - 4x + 4 - 4x + 8 + 3 {x}^{2} + 12x + 12 = 0 \\ = > 4 {x}^{2} + 4x + 24 = 0 \\ = > {x}^{2} + x + 6 = 0 \\ = > {x}^{2} + 3x - 2x + 6 = 0 \\ = > x(x + 3) - 2( x +3) = 0 \\ = > (x + 3)(x - 2) = 0 \\ = > x = - 3 \: \: and \: \: 2$