anonymous
  • anonymous
does anyone know how to do extraneous soultions for radicals?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
To 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
  • anonymous
i dont understand this at all.
anonymous
  • anonymous
Do 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.

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anonymous
  • anonymous
how do you get the solutions?
anonymous
  • anonymous
\[5\sqrt{x}=1\] is one of the problems but i dont know how to find the solution
anonymous
  • anonymous
Square 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
  • anonymous
thank you, how would i do one like this: \[(x-3)1/3=6\]
anonymous
  • anonymous
That's simple. Just multiply by 3 on both sides to get rid of the 1/3. (x-3)=6*3 x-3=18 x=21
anonymous
  • anonymous
thank 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
  • anonymous
OK, 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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