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anonymous
 5 years ago
does anyone know how to do extraneous soultions for radicals?
anonymous
 5 years ago
does anyone know how to do extraneous soultions for radicals?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To check for extraneous solutions, plug each of your solutions in for x into the radical. If you get a negative radical, the solution was introduced through squaring and can be eliminated.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont understand this at all.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have solutions like x=0, x=1, etc.? Take these and plug them into the original problem. If you get a negative under the radical, don't count that as an answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you get the solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[5\sqrt{x}=1\] is one of the problems but i dont know how to find the solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Square it, so you get 25*x=1 So x=1/25. Plug it back in 5*sqrt((1/25))=1, which works. 1/25 is not an extraneous solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you, how would i do one like this: \[(x3)1/3=6\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's simple. Just multiply by 3 on both sides to get rid of the 1/3. (x3)=6*3 x3=18 x=21

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you that helped me with another one similar to that, i have one last question on how to do one like this: \[\sqrt{x}+1=\sqrt{2x}7\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, let's get everything organized to be solved. 8=\[\sqrt{2x}\sqrt{x}\] 8=\[(\sqrt{2}1)*\sqrt{x}\] 8/\[(\sqrt{2}1)\]=sqrt(x) square both sides. then plug in your solutions to the original equation to see if they work.
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