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Mathematics 20 Online
OpenStudy (anonymous):

Solve the following equations for x. Identify any extraneous solutions. Show all work. √x+2+4=x

OpenStudy (anonymous):

I really wish you'd use parens to show what was under the radical and what wasn't..

OpenStudy (anonymous):

but there are no parens in this equation

OpenStudy (anonymous):

\[\sqrt{x+2}+4=x\] is that your question

OpenStudy (anonymous):

yes thats it

OpenStudy (anonymous):

ok do it this way \[\sqrt{x+2}=x-4\] Then square both sidez\[(\sqrt{x+2})^{2}=(x-4)^{2}\]

OpenStudy (anonymous):

\[(x+2)=x ^{2}-8x+16\]

OpenStudy (anonymous):

\[x ^{2}-9x+14=0\] \[(x-7)(x-2)=0\] solution x=7,2But 2 doesn't work..... so only sol is x=7

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