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Having difficulty understanding radicals

  1. Aug 8, 2013 #1
    I know how the very basics but then I get given a question like this.

    ##\sqrt{9-3x}## and I think I can divide both by 3. ##3\sqrt{3-x}## and so ##x=3##

    Then I get ##\sqrt{4x+12}## and again I can take 4 from each ##4\sqrt{x+3}## and so ##x=-3##

    Is this the correct way to be solving these? Because I'm looking down the page and it starts getting real complicated looking, real fast and was just wondering if there is a simple way to go about solving these?

    Simplify: ##{\sqrt\frac{9a}{8b^2}}## is the next question I face
     
    Last edited: Aug 8, 2013
  2. jcsd
  3. Aug 8, 2013 #2

    mfb

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    ##\sqrt{9-3x}## is not the same as ##3\sqrt{3-x}##. Consider x=0, for example, and see what you get in the different expressions.

    "and so x=3" does not make sense. There is no equality involved in the expressions.

    To solve anything, you need an equation. something=somethingelse.


    There are rules for square roots (more general: rules for exponents) you can use here.
     
  4. Aug 8, 2013 #3
    Also here is another one but I'm not sure it's correct:

    ##{\sqrt\frac{8}{9}}## = ##{\sqrt\frac{8}{9}\cdot{\sqrt\frac{9}{9}}}## = ##{\sqrt\frac{72}{81}}## =## \frac{\sqrt{72}}{9}##

    ?
     
  5. Aug 8, 2013 #4

    mfb

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    That is correct.
     
  6. Aug 8, 2013 #5

    symbolipoint

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    The distinction between the root symbol and grouping symbols is that although both are grouping symbols, the root symbol is a function symbol. Inside the root symbol, you can factor one or more common factors from terms, but you cannot necessarily move them outside of the root symbol; because doing so requires that you have great enough count of the factors as indicated by the root index.

    [itex]\sqrt{9-3x}[/itex] you can do [itex]\sqrt{3(3-x)}[/itex] but not further.

    ([strike]All I did was added the proper parentheses for grouping symbol balancing and the typesetting fails. [/strike] [strike]I NEED the paired parentheses in the second expression! [/strike] The typesetting did work but failed while I first made the post and mathematical notation.)
     
    Last edited: Aug 8, 2013
  7. Aug 8, 2013 #6

    symbolipoint

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    I really want the typesetting to work right but it did not so I do this instead:

    (9-3x)1/2 is the same as (3(3-x))1/2 but you cannot go any further.
     
  8. Aug 8, 2013 #7

    verty

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    This is something you will commonly see/use:

    ##\frac{1}{2} \sqrt{x} = \sqrt{\frac{1}{4}} \; \sqrt{x} = \sqrt{\frac{x}{4}}##
    ##\sqrt{4x} = \sqrt{4} \; \sqrt{x} = 2 \sqrt{x}##

    Also ##\sqrt{54} = \sqrt{9 * 6} = 3 \sqrt{6}##.

    These are all explained by replacing each radical with the corresponding exponent, like ##(\frac{x}{4})^{\frac{1}{2}}##.
     
  9. Aug 8, 2013 #8

    HallsofIvy

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    That's true but [tex]\sqrt{9}= 3[/tex] so that [tex]\frac{\sqrt{8}}{3}[/tex] would be better. In fact, since [tex]8= 4(2)[/tex] and [tex]\sqrt{8}= 2\sqrt{2}[/tex], the simplest form is [tex]\frac{2\sqrt{2}}{3}= \frac{2}{3}\sqrt{2}[/tex].
     
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