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Math Help - Radical Equation help

  1. #1
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    Radical Equation help

    Can anyone help me to set this up?

    solve: (√2m+15)=m

    everything on the left side of the equation is under the sq. root.

    thanks
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Radical Equation help

    Take the square of both sides:
    \sqrt{2m+15}=m
    \Leftrightarrow \left(\sqrt{2m+15}\right)^2=m^2

    (Don't forget to determine the 'existence conditions' because 2m+15 has to be >0 or = 0)
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  3. #3
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    Re: Radical Equation help

    ok so would the next step be -m^2+2m+15 >=0 ?

    Thanks
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Radical Equation help

    Yes, that's correct, but you have understand me wrong, I didn't say that:
    -m^2+2m+15>=0
    You just can wright:
    -m^2+2m+15=0

    I mean, that after you've calculated the solutions, you have to assure that m>=\frac{-15}{2}(conditon 1) and m>0 (condition 2) summarized m>0, because you've to assure you don't get a negative number under the square root.
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  5. #5
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    e^(i*pi)'s Avatar
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    Re: Radical Equation help

    I find it easier to take a as being positive since IMO it makes factorising it easier. To do this you can move everything over to the other side (and not forgetting to change the sign) :
    m^2-2m-15=0. This quadratic does factor but there is nothing wrong with the formula if you prefer it.

    Because you squared both sides of the equation you may have introduced extraneous solutions - solutions which work in the squared equation but not in the original one. For example we're given the equation x-4=0.
    If we square it we arrive at x^2=16 which has both 4 and -4 as "solutions" but we can see that -4 clearly doesn't work in the original equation.

    You must also consider the domain of the original equation which is what Siron pointed out by saying that 2m+15 \geq 0.

    In this example one of your solutions to the quadratic will be extraneous - you can find out which by subbing the values into the original equation
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Radical Equation help

    @Drewm:
    Have you solved this inequality now? ...
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  7. #7
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    Re: Radical Equation help

    There's NO inequality involved; just solve: m^2 - 2m - 15 = 0
    Easy to factor: (m - 5)(m + 3) = 0
    So m = 5 or m = -3; over and out!
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  8. #8
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    e^(i*pi)'s Avatar
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    Re: Radical Equation help

    Quote Originally Posted by Wilmer View Post
    There's NO inequality involved; just solve: m^2 - 2m - 15 = 0
    Easy to factor: (m - 5)(m + 3) = 0
    So m = 5 or m = -3; over and out!
    I agree with m=5

    Howevr, m=-3 is not a solution since if we put -3 into the original equation we get \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Radical Equation help

    Yes, offcourse, I was a bit confused, it's offcourse an equation.
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  10. #10
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    Re: Radical Equation help

    Quote Originally Posted by e^(i*pi) View Post
    Howevr, m=-3 is not a solution since if we put -3 into the original equation we get \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3
    Original is really +/-SQRT(2m + 15) = +/-m ?

    Anyway, IF the problem was to solve m^2 - 2m - 15 = 0 (no mention of SQRT...),
    then we wouldn't care, right?
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  11. #11
    MHF Contributor Siron's Avatar
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    Re: Radical Equation help

    Quote Originally Posted by Wilmer View Post
    Original is really +/-SQRT(2m + 15) = +/-m ?
    I don't think so, we have to reject m=-3. Because we may only square both sides if they're both positive and that's why we said m>0. And also if the equation would be:
    -\sqrt{2m+15}=-m, we would wright this as:
    m=\sqrt{2m+15}
    Quote Originally Posted by Wilmer View Post
    Anyway, IF the problem was to solve m^2 - 2m - 15 = 0 (no mention of SQRT...),
    then we wouldn't care, right?
    Right!
    Last edited by Siron; July 26th 2011 at 10:04 AM.
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