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Math Help - I need help with another problem

  1. #1
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    I need help with another problem

    The question came with the following instruction:
    Express irrational solution in simplest radical form. If the equation has no solution, write "no solution".

    (n-5)^2=20

    Thanks for your help!

    Drakmord
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  2. #2
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    Quote Originally Posted by drakmord View Post
    The question came with the following instruction:
    Express irrational solution in simplest radical form. If the equation has no solution, write "no solution".

    (n-5)^2=20

    Thanks for your help!

    Drakmord
    Think of (n-5) as "x". If you had a problem like x^2=20, what would you do to solve it? You would take the square root of both sides. Same thing here. Make sense?
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  3. #3
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    Quote Originally Posted by Jameson View Post
    Think of (n-5) as "x". If you had a problem like x^2=20, what would you do to solve it? You would take the square root of both sides. Same thing here. Make sense?
    So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

    Am I close?
    Thank you so much for your help Jameson!

    Drakmord
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by drakmord View Post
    So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

    Am I close?
    Thank you so much for your help Jameson!

    Drakmord
    Almost, but remember that when you take a square root there will be a positive root and a negative root. In other words n-5 = \pm \sqrt{20}
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  5. #5
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    Quote Originally Posted by drakmord View Post
    So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

    Am I close?
    Thank you so much for your help Jameson!

    Drakmord
    You are welcome. Remember to use the THANKS button when you find a helpful post.
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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post
    Almost, but remember that when you take a square root there will be a positive root and a negative root. In other words n-5 = \pm \sqrt{20}
    Many thanks for your help e^(i*pi)!
    So it would come out to be n-5 =2(\pm \sqrt{5})?

    drakmord
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  7. #7
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    Quote Originally Posted by drakmord View Post
    Many thanks for your help e^(i*pi)!
    So it would come out to be n-5 =2(\pm \sqrt{5})?

    drakmord
    No because that says you should either add or subtract root 5 from 2 which is not the case. The plus or minus sign should be before the two because \sqrt{20} = 2\sqrt{5} and -\sqrt{20} = -(\sqrt{20}) = -(2\sqrt{5}) = -2\sqrt{5}
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  8. #8
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    Quote Originally Posted by e^(i*pi) View Post
    No because that says you should either add or subtract root 5 from 2 which is not the case. The plus or minus sign should be before the two because \sqrt{20} = 2\sqrt{5} and -\sqrt{20} = -(\sqrt{20}) = -(2\sqrt{5}) = -2\sqrt{5}

    Ok, so the answer is: (n-5)^2=20 --> \sqrt{(n-5)^2}=\sqrt{20}--> (n-5)=\sqrt{20} --> (n-5)=\pm 2\sqrt{5} ?

    Drakmord
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  9. #9
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by drakmord View Post
    Ok, so the answer is: (n-5)^2=20 --> \sqrt{(n-5)^2}=\sqrt{20}--> (n-5)=\sqrt{20} --> (n-5)=\pm 2\sqrt{5} ?

    Drakmord
    The \pm should be in every step except the first one:

    (n-5)^2 = 20 \rightarrow \, n-5 = \pm \sqrt{20} \rightarrow \, n-5 = \pm 2\sqrt{5} \rightarrow \, n = 5 \pm 2\sqrt{5}
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  10. #10
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    Thank you for your patience and for your help e^(i*pi)!

    I think I understand it now.

    Drakmord
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  11. #11
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    Quote Originally Posted by drakmord View Post

    Thank you for your patience and for your help e^(i*pi)!

    I think I understand it now.

    Drakmord
    No problem ^.^ - it's an important point to know but confusing at first.

    It comes about because multiplying two negatives gives a positive answer
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