Solve for x ∈ R that suitable on this inequality : (√(8−x^2 ))


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Question Number 115312 by naka3546 last updated on 25/Sep/20

$S o l v e f o r x \in R t h a t s u i t a b l e o n t h i s$ $i n e q u a l i t y : \sqrt{8 - x^{2}} > x$
Commented bybemath last updated on 25/Sep/20

$i n t e r s e c t i o n t w o g r a p h$ $w e g e t 8 - x^{2} = x^{2} \Rightarrow 8 = 2 x^{2}$ $x = \pm 2 \Rightarrow s o l u t i o n s e t i s x < 2 .$
Commented byMJS_new last updated on 25/Sep/20

$you are right, two graphs$ $but f (x) = \sqrt{8 - x^{2}} is defined for - 2 \sqrt{2} ⩽ x ⩽ 2 \sqrt{2}$ $and f (x) ⩾ 0 . you are not allowed to square$
	Answered by john santu last updated on 25/Sep/20

	$(1) 8 - x^{2} ⩾ 0 \Rightarrow (\sqrt{8} + x) (\sqrt{8} - x) ⩾ 0$ $- 2 \sqrt{2} ⩽ x ⩽ 2 \sqrt{2}$ $(2) 8 - x^{2} > x^{2} \to 4 - x^{2} > 0$ $(2 + x) (2 - x) > 0$ $- 2 < x < 2$ $s o l u t i o n s e t (1) \cap (2) \Rightarrow - 2 < x < 2$
	Commented bynaka3546 last updated on 25/Sep/20

	$b u t w h y ? a n y o t h e r s o l u t i o n ?$ $x = - 2 \sqrt{2} i s r i g h t a n s w e r .$
	Commented byMJS_new last updated on 25/Sep/20

	$(2) is wrong for x < 0$
	Answered by MJS_new last updated on 25/Sep/20

	$\sqrt{8 - x^{2}} > x$ $\sqrt{8 - x^{2}} defined for - 2 \sqrt{2} ⩽ x ⩽ 2 \sqrt{2}$ $\sqrt{8 - x^{2}} ⩾ 0 \Rightarrow - 2 \sqrt{2} ⩽ x ⩽ 0 is a part of the$ $solution$ $for x > 0 we are allowed to square$ $8 - x^{2} > x^{2} \land x > 0$ $\Rightarrow 0 < x < 2$ $\Rightarrow solution is - 2 \sqrt{2} ⩽ x < 2$
	Answered by mr W last updated on 25/Sep/20

	$8 - x^{2} ⩾ 0$ $\Rightarrow - 2 \sqrt{2} ⩽ x ⩽ 2 \sqrt{2}$ $f o r - 2 \sqrt{2} ⩽ x ⩽ 0 :$ $\sqrt{8 - x^{2}} > x i s a l w a y s t r u e .$ $f o r 0 < x ⩽ 2 \sqrt{2} :$ $8 - x^{2} > x^{2}$ $\Rightarrow x < 2$ $a n s w e r :$ $- 2 \sqrt{2} ⩽ x < 2$
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