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Math Help - Radical equation problems?

  1. #1
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    Radical equation problems?

    my problem is
    square root (a+21)-1= square root(a+12) I am kind of stuck on how to solve this as I do not know what to do with -1.

    Also the next equation I need help solving is very similar
    sqrt (x+2)-7= square root (x+9) Can anyone please help me?
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  2. #2
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    \sqrt{a+21}-1=\sqrt{a+12}
    (\sqrt{a+21}-1)^2=(\sqrt{a+12})^2
    a+21 - 2\sqrt{a+21} + 1 = a+12
    Use algebra to get \sqrt{a+21} on one side. Once you do that, you can square both sides and no more square roots. Now you can solve for a. It will be a quadratic equation.

    Note: Remember once you get a solution, plug it back in to check. Especially for square root problems, you can get extra "fake" solutions. Always plug in and check.
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  3. #3
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    Quote Originally Posted by homeylova223 View Post
    my problem is
    square root (a+21)-1= square root(a+12) I am kind of stuck on how to solve this as I do not know what to do with -1.

    Also the next equation I need help solving is very similar
    sqrt (x+2)-7= square root (x+9) Can anyone please help me?
    \sqrt{a+21}-1=\sqrt{a+12}

    a+12=b

    \sqrt{b+9}-1=\sqrt{b}

    b=16
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  4. #4
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    The back of my book state the answer for the first equation is apparently 4. Also quick question how do you move -2 square root(a+2) to the other side of the equation.
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  5. #5
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    I mean -2 square root( a+21) in the work snowtea did.
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  6. #6
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    How do you get anything on one side?

    Pretend \sqrt{a+21} is just x.

    I.e. a+21 - 2x + 1 = a+12

    How do you solve for x?
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  7. #7
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    would you just add -2 square root(a+21) to the other side of the equation?
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  8. #8
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    Yes, can you solve for x in a+21 - 2x + 1 = a+12?
    You should get x = ?, then just replace it with \sqrt{a+21} = ?
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  9. #9
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    Yes.
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  10. #10
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    Quote Originally Posted by homeylova223 View Post
    would you just add -2 square root(a+21) to the other side of the equation?
    No, you add 2\sqrt{a+21} to both sides.

    You cancel -2\sqrt{a+21} from the side it's on.

    If two sides are equal, then if you add the same thing to both sides, they will still be equal.

    As in.... 5=5, 5+2=5+2 or 5-2=5-2 etc

    therefore

    a+22-2\sqrt{a+21}=a+12

    Add 2\sqrt{a+21} to both sides and then subtract (a+12) from both sides.

    Alternatively, subtract (a+22) from both sides.
    Then divide both sides by -2 and square both sides.


    You could also go this route...

    \displaystyle\sqrt{a+21}-1=\sqrt{a+12}

    \displaystyle\sqrt{\left(\sqrt{a+21}-1\right)^2}=\sqrt{a+12}

    \displaystyle\sqrt{a+21-2\sqrt{a+21}+1}=\sqrt{a+12}

    \sqrt{a+22-2\sqrt{a+21}}=\sqrt{a+12}

    a+22-2\sqrt{a+21}=a+12

    a+12+10-2\sqrt{a+21}=a+12

    which means that

    10-2\sqrt{a+21}=0

    If two values are equal, when we subtract them the answer is zero...

    10=2\sqrt{a+21}\Rightarrow\sqrt{a+21}=5\Rightarrow  \ a+21=25
    Last edited by Archie Meade; December 16th 2010 at 06:39 AM. Reason: alternative method
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  11. #11
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    So far this is how far I got

    I have
    a^2+21-2 square root (a+21)+1= a+12

    The I just turn sqrt (a+21) into A
    So I get
    A^2+21-2(A)+1= a+12

    A^2+21+1= 3A+12
    Then I would get
    A^2-3A+10 I am stuck
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  12. #12
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    Take another look at Archie Meads post. It pretty much works out the entire problem.
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  13. #13
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    Quote Originally Posted by homeylova223 View Post
    So far this is how far I got

    I have
    a^2+21-2 square root (a+21)+1= a+12 there is a typo here on a^2

    The I just turn sqrt (a+21) into A
    So I get
    A^2+21-2(A)+1= a+12 another typo on A^2

    A^2+21+1= 3A+12 oops! "A" is not equal to "a"

    Then I would get
    A^2-3A+10 I am stuck


    If A=\sqrt{a+21}\Rightarrow\ A^2=a+21\Rightarrow\ a+12=A^2-9

    Then....

    \left[\sqrt{a+21}\right]^2-2\sqrt{a+21}+1=a+12

    A=\sqrt{a+21}

    A^2-2A+1=A^2-9

    -(2A-1)=-9

    2A-1=9

    A=5\Rightarrow\sqrt{a+21}=\sqrt{25}\Rightarrow\ a+21=25
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