Respuesta :

caylus
Hello,

[tex] \sqrt{2x+4}- \sqrt{x} =2\\ ==\ \textgreater \ \sqrt{2x+4}=2+ \sqrt{x}\\ ==\ \textgreater \ 2x+4=4+4 \sqrt{x} +x \\ ==\ \textgreater \ x=4 \sqrt{x} \\ ==\ \textgreater \ x^2=16*x\\ ==\ \textgreater \ x^2-16x=0\\ ==\ \textgreater \ x(x-16)=0\\ ==\ \textgreater \ x=0 \ or \ x=16\\ Answer B [/tex]
your answer is x+0 and x=-16

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