Question

What is the solution

Answer 1

Answer:

Step-by-step explanation:

$\sqrt{2x+4} -\sqrt{x} =2\\\sqrt{2x+4} =2+\sqrt{x} \\squaring\\2x+4=4+x+4\sqrt{x} \\2x+4-4-x=4\sqrt{x} \\x=4\sqrt{x} \\again~squaring\\x^2=16x\\x^2-16x=0\\x(x-16)=0\\x=0,16$