Find the all valid value of x x^2 − (√(4+x)) = 4


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Question Number 172282 by BaliramSingh last updated on 25/Jun/22

$F i n d t h e a l l v a l i d v a l u e o f x$ $x^{2} - \sqrt{4 + x} = 4$
	Answered by dumitrel last updated on 25/Jun/22

	$x^{2} - 4 ⩾ 0; x + 4 ⩾ 0$ $x^{2} - 4 = \sqrt{4 + x} = y$ $x^{2} - y = 4$ $y^{2} - x = 4 \Rightarrow x^{2} - y^{2} + (x - y) = 0$ $(x - y) (x + y + 1) = 0$ $x = y \Rightarrow x^{2} - x - 4 = 0 \Rightarrow x = \frac{1 + \sqrt{17}}{2}$ $x + y + 1 = 0 \Rightarrow x^{2} + x - 3 = 0$ $x = \frac{- 1 - \sqrt{13}}{2}$
	Commented by BaliramSingh last updated on 25/Jun/22

	$T h a n k s s i r$
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