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anonymous

  • 5 years ago

does anyone know how to do extraneous soultions for radicals?

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  1. anonymous
    • 5 years ago
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    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.

  2. anonymous
    • 5 years ago
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    i dont understand this at all.

  3. anonymous
    • 5 years ago
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    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.

  4. anonymous
    • 5 years ago
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    how do you get the solutions?

  5. anonymous
    • 5 years ago
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    \[5\sqrt{x}=1\] is one of the problems but i dont know how to find the solution

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    thank you, how would i do one like this: \[(x-3)1/3=6\]

  8. anonymous
    • 5 years ago
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    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

  9. anonymous
    • 5 years ago
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    thank you that helped me with another one similar to that, i have one last question on how to do one like this: \[\sqrt{x}+1=\sqrt{2x}-7\]

  10. anonymous
    • 5 years ago
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    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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