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Thread: 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:
    $\displaystyle \sqrt{2m+15}=m$
    $\displaystyle \Leftrightarrow \left(\sqrt{2m+15}\right)^2=m^2$

    (Don't forget to determine the 'existence conditions' because $\displaystyle 2m+15$ has to be $\displaystyle >0$ or $\displaystyle =$ 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:
    $\displaystyle -m^2+2m+15>=0$
    You just can wright:
    $\displaystyle -m^2+2m+15=0$

    I mean, that after you've calculated the solutions, you have to assure that $\displaystyle m>=\frac{-15}{2}$(conditon 1) and $\displaystyle m>0$ (condition 2) summarized $\displaystyle m>0$, 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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    Re: Radical Equation help

    I find it easier to take $\displaystyle a$ 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) :
    $\displaystyle m^2-2m-15=0$. 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 $\displaystyle x-4=0$.
    If we square it we arrive at $\displaystyle x^2=16$ 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 $\displaystyle 2m+15 \geq 0$.

    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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    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 $\displaystyle m=5$

    Howevr, $\displaystyle m=-3$ is not a solution since if we put -3 into the original equation we get $\displaystyle \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3$
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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, $\displaystyle m=-3$ is not a solution since if we put -3 into the original equation we get $\displaystyle \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3$
    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 $\displaystyle m=-3$. Because we may only square both sides if they're both positive and that's why we said $\displaystyle m>0$. And also if the equation would be:
    $\displaystyle -\sqrt{2m+15}=-m$, we would wright this as:
    $\displaystyle m=\sqrt{2m+15}$
    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; Jul 26th 2011 at 09:04 AM.
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