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Thread: Radical Equation Help

  1. #1
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    Radical Equation Help

    $\displaystyle
    3\sqrt{x} = \sqrt{x+9}
    $

    Evaluating this i can see that both sides of the equation are equal if we de simplify the left side or simplify the right side since 9 is a perfect square....

    But problem is i originally attempted to solve this problem by squaring both sides to get rid of the sqrt leaving me with:

    $\displaystyle
    9x = x+9
    $

    I essentially distributed the exponent to the 3 and the sqrt of x individually. Am I not able to do this must i treat them as one somehow?

    This is not the same as we now have an inequality. Can someone explain to me what i did wrong... I know it is simpler to just simplify since its a perfect square, but i am not satisfied doin things one way math should always work no matter which way we try to solve an equation.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by nmbala View Post
    $\displaystyle
    3\sqrt{x} = \sqrt{x+9}
    $

    Evaluating this i can see that both sides of the equation are equal if we de simplify the left side or simplify the right side since 9 is a perfect square....

    By problem is i originally attempted to solve this problem by squaring both sides to get rid of the sqrt leaving me with:

    $\displaystyle
    9x = x+9
    $

    This is not the same as we now have an inequality. Can someone explain to me what i did wrong... I know it is simpler to just simplify since its a perfect square, but i am not satisfied doin things one way math should always work no matter which way we try to solve an equation.
    Hi nmbala,

    Where do you have an inequality?

    $\displaystyle
    9x=x+9$

    $\displaystyle 8x=9$

    $\displaystyle x=\frac{9}{8}$
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  3. #3
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    Hmm i was doin this which must be where i am confused.

    $\displaystyle
    9x=x+9
    $

    $\displaystyle
    \frac{9x}{9} = \frac {x+9}{9}
    $

    $\displaystyle
    1x = 1x +9
    $
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by nmbala View Post
    Hmm i was doin this which must be where i am confused.

    $\displaystyle
    9x=x+9
    $

    $\displaystyle
    \frac{9x}{9} = \frac {x+9}{9}
    $

    $\displaystyle
    1x = 1x +9
    $
    Your last step does not follow from the previous one.

    $\displaystyle \frac{9x}{9}=\frac{x+9}{9}$

    $\displaystyle 1x=\frac{x+9}{9}$

    $\displaystyle x=\frac{x}{9}+1$

    $\displaystyle x-\frac{x}{9}=1$

    $\displaystyle \frac{8}{9}x=1$

    $\displaystyle x=\frac{1}{\frac{8}{9}}$

    $\displaystyle x=\frac{9}{8}$

    But I would never go about it that way......too long and unnecessary.
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  5. #5
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    Opps was suppose to be +1 not 9... Regardless i see that i need to carry that 9 to the x also. Thanks for your help!

    Last question regarding steps 4 and 5 of your previous post. What is the logic behind:
    $\displaystyle x-\frac{x}{9}=1$

    $\displaystyle \frac {8}{9}x =1$

    Appears I have forgotten something else from class 3 years ago...
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  6. #6
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    Smile

    Hello
    $\displaystyle x-\frac{x}{9}=1\Leftrightarrow \frac{9x-x}{9}=1\Leftrightarrow \frac{8x}{9}=1$
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  7. #7
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    $\displaystyle x - \frac{x}{9} = \frac{9}{9}x - \frac{1}{9}x = \left( \frac{9}{9} - \frac{1}{9} \right)x = \frac{8}{9}x$
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  8. #8
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    Quote Originally Posted by Raoh View Post
    Hello
    $\displaystyle x-\frac{x}{9}=1\Leftrightarrow \frac{9x-x}{9}=1\Leftrightarrow \frac{8x}{9}=1$

    Right Lowest common denominator to add unlike fractions. Thanks!

    I need to start remembering to look at things as $\displaystyle \frac{x}{1} - \frac{x}{9}$
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