Solve for x (√(2x+3)) −(√(x−2)) =(√(x+2))


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Question Number 218093 by ArshadS last updated on 29/Mar/25

$S o l v e f o r x$ $\sqrt{2 x + 3} - \sqrt{x - 2} = \sqrt{x + 2}$
	Answered by vnm last updated on 29/Mar/25

	$2 x + 3 = {(\sqrt{x + 2} + \sqrt{x - 2})}^{2} =$ $x + 2 + x - 2 + 2 \sqrt{(x + 2) (x - 2)}$ $3 = 2 \sqrt{x^{2} - 4}$ $x^{2} = {(\frac{3}{2})}^{2} + 4$ $x = \frac{5}{2}$ $- \frac{5}{2} {c a n}^{'} t b e t h e s o l u t i o n$
	Commented by ArshadS last updated on 29/Mar/25

	$T h a n k y o u s i r$
	Answered by Marzuk last updated on 30/Mar/25

	$\sqrt{2 x + 3} - \sqrt{x - 2} = \sqrt{x + 2}$ $o r, \sqrt{2 x + 3} = \sqrt{x + 2} + \sqrt{x - 2}$ $o r, 2 x + 3 = x + 2 + x - 2 + 2 \sqrt{x + 2} . \sqrt{x - 2} [s q u a r i n g e a c h s i d e]$ $o r, 2 x + 3 = 2 x + 2 \sqrt{x + 2} . \sqrt{x - 2}$ $o r, \sqrt{x^{2} - 4} = \frac{3}{2}$ $o r, x^{2} = {(\frac{3}{2})}^{2} + 4$ $o r, x^{2} = \frac{25}{4}$ $o r, x = \sqrt{\frac{25}{4}}$ $∴ x = \frac{5}{2}$ $I {d i d n}^{'} t s h o w s o m e s t e p s h e r e .$
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