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Math Help - square roots

  1. #1
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    square roots

    1.the square root of y+4 -2=3

    2.the cubed root of 6x+9 +5=2
    the space implies that the other numbers are not under the square root
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  2. #2
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    any help would be greatly appreciated
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  3. #3
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    Quote Originally Posted by imbadatmath View Post
    1. \sqrt{y+4} - 2 = 3
    what is wrong with \sqrt{y+4} = 5
    y+4 = 25
    y = 21

    Or am i reading you question wrong ?

    Bobak
    Last edited by bobak; April 20th 2008 at 03:43 PM.
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  4. #4
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    <br />
\sqrt{y+4}-2 = 3<br />
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  5. #5
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    this is how it was suposed to read
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  6. #6
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    Quote Originally Posted by imbadatmath View Post
    this is how it was suposed to read
    Your original equation is: ....... \sqrt{y+4}-3=2 ....... Add 3 on both sides.

    You'll get bobak's equation.

    But since squaring both sides of an equation is not an equivalent operation you must prove if the solution is valid or not. Plug in the solution into the original equation and prove if the equation is true:

    \sqrt{21+4}-3 \ \buildrel {\rm?} \over {\rm=} \ 2

    5-3 \buildrel {\rm?} \over {\rm=} \ 2

    2 \ \buildrel {\rm!} \over {\rm=} \ 2....... So y = 21 is a valid solution.

    to #2:

    \sqrt[3]{6x+9} + 5 = 2~\iff~ \sqrt[3]{6x+9}=-3

    Cube both sides. You'll get:

    6x+9 = -27 ~\implies~ 6x =-36~\implies~x = -6

    Plug in this value into the original equation:

    \sqrt[3]{6 \cdot (-6)+9}+5 \buildrel {\rm?} \over {\rm=} 2~\iff~\sqrt[3]{-27}+5 \buildrel {\rm?} \over {\rm=} 2~\iff~ -3+5 \buildrel {\rm?} \over {\rm=} 2....... which is obviously true.

    And therefore x = -6 is a solution of this equation.
    Last edited by earboth; April 23rd 2008 at 12:15 AM. Reason: found a typo
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