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Math Help - [SOLVED] Expressing in the form?

  1. #1
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    [SOLVED] Expressing in the form?

    Express \sqrt{44-24\sqrt{2}} in the form a + b\sqrt{2}
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  2. #2
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    Hello, Mark!

    Express \sqrt{44-24\sqrt{2}} in the form a + b\sqrt{2}
    \text{We have: }\;\sqrt{44-24\sqrt{2}} \;=\;a + b\sqrt{2}\;\;\text{ where }a\text{ and }b\text{ are rational numbers.}

    Square both sides: . 44 - 24\sqrt{2} \:=\:(a+b\sqrt{2})^2 \:=\:a^2 + 2ab\sqrt{2} + 2b^2

    And we have: . (a^2 + 2b^2) + (2ab)\sqrt{2} \;=\;44-24\sqrt{2}


    Equate coefficients: . \begin{array}{ccc}a^2+b^2 \:=\:44 & {\color{blue}[1]} \\ 2ab \:=\:-24 & {\color{blue}[2]} \end{array}

    From {\color{blue}[2]}\!:\;\;b \:=\:-\frac{12}{a}\;\;{\color{blue}[3]}

    \text{Substitute into }{\color{blue}[1]}\!:\;\;a^2 + 2\left(-\frac{12}{a}\right)^2 \:=\:44  \quad\Rightarrow\quad a^2 + \frac{288}{a^2} \:=\:44<br />

    Multiply by a^2\!:\;\;a^4 + 288 \:=\:44a^2 \quad\Rightarrow\quad a^4 - 44a^2 + 288 \:-\:0

    Factor: . (a^2 - 36)(a^2-8) \:=\:0 \quad\Rightarrow\quad (a-6)(a+6)(a^2-8) \:=\:0

    . . And the rational roots are: . a \:=\:\pm 6

    Substitute into {\color{blue}[3]}\!:\;\;b \:=\:-\frac{12}{\pm6} \:=\:\mp 2


    There are two solutions: . a + b\sqrt{2}\;=\;\begin{Bmatrix}6 - 2\sqrt{2} \\ \text{-}6 + 2\sqrt{2} \end{Bmatrix}

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  3. #3
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    What's the difference between \pm and \mp? Why do you have to inverse them to get \mp2? Isn't it okay to just put \pm2?
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  4. #4
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    Quote Originally Posted by mark1950 View Post
    What's the difference between \pm and \mp? Why do you have to inverse them to get \mp2? Isn't it okay to just put \pm2?
    No. Looking at this from Soroban's post:
    b \:=\:-\frac{12}{\pm6} \:=\:\mp 2

    See the \pm6 in the denominator? The \mp sign indicates that the number that follows has to be the opposite of whatever sign the number after the \pm is. If the denominator is +6, then it simplifies to -2, and if the denominator is -6, it simplifies to +2.

    The \mp isn't used as often as \pm. If you know trig, then you may have seen the sum/difference identity for cosine combined into one identity like this:
    \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B.

    Of course, the two identities are these:
    \cos(A + B) = \cos A \cos B - \sin A \sin B.
    \cos(A - B) = \cos A \cos B + \sin A \sin B.

    Note how the sign changes on the right side. That's the reason we use the \mp sign.


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