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Math Help - Modulus and Irrational Equations

  1. #1
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    Modulus and Irrational Equations

    Hello all my name is larsson and I am 30 years of age and am trying to teach myself Maths. I was able to do most of the questions in this section but the last few have me stumped

    (the square root of x) + 1 = ( the square root of x + 9 )

    It says these questions need squaring twice and the answer is 16 but I can't work it out.

    Any help would be great
    Thanks and regards
    Larsson
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  2. #2
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     \sqrt{x} +1 = \sqrt{x+9}

    square both sides

     x + 2 \sqrt{x} +1 = x+9
     2 \sqrt{x}= 8
     \sqrt{x}= 4
    x = 16


    With equations the involve squaring, it is important to check your solutions. This is because the process of squaring an equations is not reversible.

    From going to your first step to your second step you introduce an additional equation into the problem.

    The expression x + 2 \sqrt{x} +1 = x+9 could have either come form squaring this \sqrt{x} +1 = \sqrt{x+9} or - \sqrt{x} - 1 = \sqrt{x+9}, so the solutions to obtain through this process could be solutions to either equation. additional solutions obtained form squaring an equation are called extraneous roots. In this case you don't have the problem but watch out for them.

    Bobak
    Last edited by bobak; April 13th 2008 at 12:48 PM.
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  3. #3
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    Thank you Bobak for taking the time to reply to me, just one thing I don't understand - how to get the second line of the solution

    x + 2(square root of x ) + 1 = x + 9
    Its the red part I don't understand

    Thanks a million
    larsson


    square both sides

     x + 2 \sqrt{x} +1 = x+9
     2 \sqrt{x}= 8
     \sqrt{x}= 4
    x = 16


    With equations the involve squaring, it is important to check your solutions. This is because the process of squaring an equations is not reversible.

    From going to your first step to your second step you introduce an additional equation into the problem.

    The expression x + 2 \Sqrt{x} +1 = x+9 could have either come form squaring this \Sqrt{x} +1 = \Sqrt{x+9} or - \Sqrt{x} - 1 = \Sqrt{x+9}, so the solutions to obtain through this process could be solutions to either equation. additional solutions obtained form squaring an equation are called extraneous roots. In this case you don't have the problem but watch out for them.

    Bobak[/quote]
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  4. #4
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    Hello larsson

    Sorry for the late reply

    I squared both sides.

    we started with \sqrt{x} +1 = \sqrt{x+9}<br />

    then

    (\sqrt{x} +1)^2 = (\sqrt{x+9})^2<br />

    (\sqrt{x})^2 + 2 \sqrt{x} +1 = x+9

    x + 2 \sqrt{x} +1 = x+9

    Is that more clear ?

    Do you understand why this step is not reversible ?
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  5. #5
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    Just to elaborate a tad more:

    (\sqrt{x} +1)^2 = (\sqrt{x} + 1) \cdot (\sqrt{x} + 1)
     = \sqrt{x} \cdot \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \cdot 1 + 1 \cdot 1
    = (\sqrt{x})^2 + 2 \cdot \sqrt{x} + 1 = x + 2\sqrt{x} + 1
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  6. #6
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    Thanks

    Thanks alot to Bobak and Flippy for taking the time to reply to me. I understand now how to do these equations.

    Once again thank you both for spending time to help me understand!!!!

    Regards
    Larsson
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