Question 1042927
{{{sqrt(2x)+4=x}}}



{{{sqrt(2x)+4-4=x-4}}} Subtract 4 from both sides



{{{sqrt(2x)=x-4}}}



{{{(sqrt(2x))^2=(x-4)^2}}} Square both sides to undo the square root



{{{2x=x^2-8x+16}}}



{{{2x-2x=x^2-8x+16-2x}}}



{{{0=x^2-10x+16}}}



{{{x^2-10x+16=0}}}



{{{(x-8)(x-2)=0}}}



{{{x-8=0}}} or {{{x-2=0}}}



{{{x=8}}} or {{{x=2}}}



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The two possible solutions are {{{x=8}}} or {{{x=2}}}



Let's check {{{x=8}}}



{{{sqrt(2x)+4=x}}}



{{{sqrt(2*8)+4=8}}} Replace every x with 8



{{{sqrt(16)+4=8}}}



{{{4+4=8}}}



{{{8=8}}} Possible solution is confirmed. So {{{x=8}}} is a true solution



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Let's check {{{x=2}}}



{{{sqrt(2x)+4=x}}}



{{{sqrt(2*2)+4=2}}} Replace every x with 2



{{{sqrt(4)+4=2}}}



{{{2+4=2}}}



{{{6=2}}} This is false, so {{{x = 2}}} is extraneous. It is NOT a true solution.



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Final Answer:



{{{x=8}}} is the only true solution to the original equation.