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Math Help - Surd

  1. #1
    Senior Member slevvio's Avatar
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    Surd

    How do I simplify \sqrt{9 + 4\sqrt{2}} to 1+2\sqrt{2}?

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  2. #2
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    Hello, slevvio!

    \text{How do I simplify }\,\sqrt{9 + 4\sqrt{2}}\,\text{ to }\,1+2\sqrt{2}\;?
    I don't think there is an easy way . . .

    First, we must suspect that 9 + 4\sqrt{2} is a square.

    Then we let: . \sqrt{9 + 4\sqrt{2}} \;=\;a + b\sqrt{2}
    . . where both a\text{ and }b are rational numbers.


    Square both sides: . 9 + 4\sqrt{2} \;=\;\left(a + b\sqrt{2}\right)^2

    . . and we have:. . a^2 + 2b^2) + 2ab\sqrt{2}" alt="9 + 4\sqrt{2} \;=\a^2 + 2b^2) + 2ab\sqrt{2}" />


    These two numbers are equal if their corresponding coefficients are equal.

    So we have: . \begin{Bmatrix}a^2+2b^2 & = & 9 \\ 2ab &=&4\end{Bmatrix}

    The second equation gives us: . b \:=\:\frac{2}{a}

    Substitute into the first equation: . a^2 + 2\left(\frac{2}{a}\right)^2 \:=\:9 \quad\Rightarrow\quad a^2 + \frac{8}{a^2} \:=\:9

    Multiply by a^2\!:\;\;a^4 + 8 \:=\:9a^2\quad\Rightarrow\quad a^4 - 9a^2 + 8 \:=\:0

    . . which factors: . (a^2-1)(a^2-8) \:=\:0

    . . and has roots: . a \;=\;\pm1,\:\pm2\sqrt{2}

    Since a must be rational and the original square root is positive,
    . . the only acceptable root is: .  a\,=\,1\quad\Rightarrow\quad b \,=\,2


    Therefore: . \sqrt{9 + 4\sqrt{2}} \;=\;1 + 2\sqrt{2}

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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Soroban View Post
    I don't think there is an easy way . . .
    Of course there is one more dear Soroban

    Quote Originally Posted by slevvio View Post
    How do I simplify \sqrt{9 + 4\sqrt{2}} to 1+2\sqrt{2}?
    \sqrt {9 + 4\sqrt 2 }  = \sqrt {\left( {1 + 4\sqrt 2  + 8} \right)}  = \sqrt {\left( {1 + 2\sqrt 2 } \right)^2 }  = \left| {1 + 2\sqrt 2 } \right|.

    Since the quantity inside bars is positive, we have that \sqrt {9 + 4\sqrt 2 }  = 1 + 2\sqrt 2 .
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  4. #4
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    Hello, Krizalid!


    Of course, you are right . . . That is a valid method.

    I use that method when I'm quite certain that we have a square.


    I've done it enough times that I suspect that 8 + 2\sqrt{15} is a square
    . . because: . 8 \:= \:3 + 5\;\hdots\;15 \:=\:3\cdot5\;\hdots\;\text{ and that coefficient is 2.}

    And, sure enough: . 8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2

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