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Math Help - Simplification

  1. #1
    Senior Member DivideBy0's Avatar
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    Simplification

    Have a go at this:

    sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}

    How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!
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  2. #2
    Bar0n janvdl's Avatar
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    Keep squaring on both sides until the roots are gone
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  3. #3
    Senior Member DivideBy0's Avatar
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    Don't see how that's very relevant... I'm not solving for a variable.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by DivideBy0 View Post
    Have a go at this:

    sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}

    How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!
    Square both sides:

    the left hand side is:

    7 - 4 sqrt(3),

    and the rigth hand side is:

    (2 - sqrt(3))(2-sqrt(3)) = 4 -4 sqrt(3) + 3 = 7 - 4 sqrt(3).

    Now we need the tan 1/2 angle formula:

    tan(A/2) = sqrt((1-cos(A))/(1+cos(A))

    Now put A=30 degrees, then cos(30)=sqrt(3)/2, SO:

    tan(15) = sqrt((1-sqrt(3)/2)/(1+sqrt(3)/2)) = sqrt((2-sqrt(3)/(2+sqrt(3))

    but:

    (2-sqrt(3)/(2+sqrt(3) = (2-sqrt(3))^2/[(2+sqrt(3))(2-sqrt(3))]

    .............................= (4 -4 sqrt(3) +3)/(4-3) = 7-sqrt(3)

    So:

    tan(15) = sqrt(7 -sqrt(3)) = 2 - sqrt(3)

    RonL
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  5. #5
    Senior Member DivideBy0's Avatar
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    If you just had sqrt{7-4sqrt{3}}, with no other information, or knowledge that it was a half-angle, would it be possible to simplify?
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  6. #6
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    Hello, DivideBy0!

    Suppose we don't know the RHS . . .


    Find: .sqrt[7 - 4sqrt{3}]
    Here is a primitive method for finding the square root.

    . . . . . . . . . _ - . . ._______
    Let: .a + b√3 .= .√7 - 4√3 . where a and b are integers
    . . . . . . . . . . . . . . . . . . . ._ . . . . . ._______
    Square both sides: .(a + b√3) .= .(√7 - 4√3)
    . . . . . . . . . . . . . . . . . ._ . . . . . . - . - . . ._
    and we have: .a + 2ab√3 + 3b .= .7 - 4√3
    . . . . . . . . . . . . . . . . . . . _ . . . . . . . ._
    Then: .(a + 3b) + (2ab)√3 .= .7 - 4√3


    Equate coefficients: . a + 3b .= .7 .[1]
    . . . . . . . . . . . . . . . . . 2ab . . = -4 .[2]

    And solve the system of equations . . .

    From [2], we have: .b = -2/a .[3]

    Substitute into [1]: .a + 3(-2/a) .= .7

    . . which simplifies to: .a^4 - 7a + 12 .= .0

    . . which factors: .(a - 3)(a - 4) .= .0
    . . . . . . . . . . . . . . . . . . . _
    . . and has roots: .a .= .√3, 2


    Since a is an integer, we will use: .a = 2

    Substitute into [3]: .b = -2/2 = -1

    . . . . . . . . . . . _______ . . . . . . . _
    . . Therefore: .√7 - 4√3 . = . 2 - √3

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  7. #7
    Senior Member DivideBy0's Avatar
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    Hmm interesting, thanks for that
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  8. #8
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    Quote Originally Posted by DivideBy0 View Post
    Hmm interesting, thanks for that
    Note, Soroban's method does not always work.
    This is mine 53th Post!!!
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  9. #9
    Senior Member DivideBy0's Avatar
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    Just a quickie, for Soroban's method, why was +2 chosen over -2? How did you know it was the right one to use?
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  10. #10
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    Hello, DivideBy0!

    Why was +2 chosen over -2? How did you know it was the right one to use? .I didn't
    It was laziness on my part . . . sorry.


    Choosing a = +2, we get: b = -1, _
    . . . and the square root is: . 2 - √3

    Choosing a = -2, we get: b = 1, - -_
    . . . and the square root is: .-2 + √3


    As you suspected, there are two square roots for any quantity (except 0):
    . . . . _______ . . . . . . . ._
    . . . √7 - 4√3 .= .(2 - √3)

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