Question #c2658

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Question #c2658

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Answer 1 · 2018-01-27T16:21:23

$x = (1 + \sqrt{5}), (1 - \sqrt{5})$ $x = (1+sqrt5), (1-sqrt5)$

Explanation:

Given, $x^{2} - 2 x = 4$ $x^2-2x=4$
$\Rightarrow x^{2} - 2 x - 4 = 0$ $rArr x^2-2x-4=0$
$\Rightarrow x^{2} - 2 x + 1 - 5 = 0$ $rArr x^2-2x+1-5=0$
$\Rightarrow {(x - 1)}^{2} - {(\sqrt{5})}^{2} = 0$ $rArr (x-1)^2-(sqrt5)^2=0$
$\Rightarrow [(x - 1) + \sqrt{5}] [(x - 1) - \sqrt{5}] = 0$ $rArr [(x-1)+sqrt5][(x-1)-sqrt5]=0$
$\Rightarrow [(x - 1) + \sqrt{5}] = 0, [(x - 1) - \sqrt{5}] = 0$ $rArr [(x-1)+sqrt5]=0,[(x-1)-sqrt5]=0$
$\Rightarrow (x - 1 + \sqrt{5}) = 0, (x - 1 - \sqrt{5}) = 0$ $rArr (x-1+sqrt5)=0, (x-1-sqrt5)=0$
$\Rightarrow x = (1 - \sqrt{5}), (1 + \sqrt{5})$ $rArr x = (1-sqrt5), (1+sqrt5)$

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Question #c2658

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