sqrt(2\sqrt{x+2)}=sqrt(x+2)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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√2√x+2=√x+2

Solution

x=2, x=−2

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√2√x+2=√x+2

Remove square roots

4x+8=x²+4x+4

Solve 4x+8=x²+4x+4: x=2, x=−2

x=2, x=−2

Verify Solutions: x=2 True, x=−2 True

The solutions are

x=2, x=−2

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

sqrt(2\sqrt{x+2)}=sqrt(x+2)

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