Question

x-2/x+2+x+2/x-2=4 solve the quadratic equation the answer is 2 root 3, - 2 root 3 please answer if you know it​

Answer 1

Question :-

$\frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = 4$

Solution :-

$= > \frac{(x - 2)(x - 2) + (x + 2)(x + 2)}{(x + 2)(x - 2)} = 4 \\ = > \frac{ {x}^{2} - 2x - 2x + 4 + {x + 2x + 2x + 4} }{ {x}^{2} - 2x + 2x - 4 } = 4 \\ = > \frac{ {2x}^{2} + 8 }{ {x}^{2} - 4 } = 4 \\ = > {2x}^{2} + 8 = {4x}^{2} - 16 \\ = > {2x}^{2} = 24 \\ = > {x}^{2} = 12 \\ = > x = + 2 \sqrt{3} \: or \: x \: = - 2 \sqrt{3}$

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Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

The equation is $\sf \fbox{ \frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = 4 }$

On simplifying it , we get

$\sf \hookrightarrow \frac{{(x - 2)}^{2} + {(x + 2)}^{2} }{ {(x)}^{2} - {(2)}^{2} } = 4 \\ \\ \sf \hookrightarrow {(x)}^{2} + {(2)}^{2} - 4x + {(x)}^{2} + {(2)}^{2} + 4x = 4 {(x)}^{2} - 16 \\ \\ \sf \hookrightarrow 2 {(x)}^{2} + 8 = 4 {(x)}^{2} - 16 \\ \\ \sf \hookrightarrow - 2 {(x)}^{2} = - 24 \\ \\\sf \hookrightarrow {(x)}^{2} = \frac{ - 24}{ - 2} \\ \\ \sf \hookrightarrow {(x)}^{2} = \sqrt{12} \\ \\ \sf \hookrightarrow x = + \sqrt{12} \: \: or \: \: x = - \sqrt{12}$

It can be written as ,

$\sf \hookrightarrow x = + 2 \sqrt{3} \: \: or \: \: x = - 2 \sqrt{3}$

Hence , roots of given quadratic equation are +2√3 and -2√3

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