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Solve the following equations for x. Identify any extraneous solutions. Show all work. √x+2+4=x
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I really wish you'd use parens to show what was under the radical and what wasn't..
but there are no parens in this equation
\[\sqrt{x+2}+4=x\] is that your question
yes thats it
ok do it this way \[\sqrt{x+2}=x-4\] Then square both sidez\[(\sqrt{x+2})^{2}=(x-4)^{2}\]
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\[(x+2)=x ^{2}-8x+16\]
\[x ^{2}-9x+14=0\] \[(x-7)(x-2)=0\] solution x=7,2But 2 doesn't work..... so only sol is x=7
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