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Math Help - Simplify the equations

  1. #1
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    Simplify the equations

    Hi
    Need help on the following equations:
    Express the following in the form x+y\sqrt2 with x and y rational numbers:
    1) (7+5\sqrt2)^{\frac{1}{3}}

    Express the following numbers in the form x+y\sqrt{n} where x and y are rational numbers and n is an integer:
    2) \frac{1}{7+5\sqrt2}
    This is what i have done and i looked in the book's answers but i don't understand how they got the answer to be -7+5\sqrt5:
    =\frac{1(7+5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}

    =\frac{(7+5\sqrt2)}{(7-5\sqrt2)}

    P.S
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  2. #2
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    Not sure about the first one. But I think I got the second one for you.

    \frac{(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}

    \frac{(7-5\sqrt2)}{(7)^2-(5\sqrt2)^2}

    \frac{(7-5\sqrt2)}{49-50}

    \frac{(7-5\sqrt2)}{-1}

    -7+5\sqrt2

    Basically, you multiply and divide by the same thing. In the case of making the denominator (lower part of the fraction) simple, you multiply both nominator and denominator with (7-5\sqrt{2})

    Edit: Wait, you sure the book says \sqrt5 for the answer?
    Last edited by dkaksl; January 2nd 2010 at 03:12 PM.
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  3. #3
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    (7+5\sqrt2)^{\frac{1}{3}}

    Let  x + y\sqrt2 =  ^3\sqrt{7 + 5 \sqrt 2}

     (x + y\sqrt2)^3 = 7 + 5 \sqrt 2

    Expand the left hand side and collect your terms

    __________________________________________________

    Express the following numbers in the form x+y\sqrt{n} where x and y are rational numbers and n is an integer:

    The answer given by your book is wrong (or you made a typo)

    \frac{1}{7+5\sqrt2}

    =\frac{1(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)} Expanding the denominator

    =\frac{(7-5\sqrt2)}{(49+ 5\sqrt 2 - 5\sqrt 2 - 50)}

     = \frac{(7-5\sqrt2)}{-1} = -1(7-5\sqrt2)

     = -7 + 5\sqrt2
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  4. #4
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    thanks for the solutions.
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  5. #5
    Senior Member Stroodle's Avatar
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    For the first question:

    x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}

    \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2}

    \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}

    So from this you can now get two equations to solve simultaneously:

    x^3+6xy^2=7

    and 2y^3+3x^2y=5

    Hence x=1 and y=1

    \Rightarrow 1+\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}
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  6. #6
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    Quote Originally Posted by Gusbob View Post
    (7+5\sqrt2)^{\frac{1}{3}}

    Let  x + y\sqrt2 =  ^3\sqrt{7 + 5 \sqrt 2}

     (x + y\sqrt2)^3 = 7 + 5 \sqrt 2

    Expand the left hand side and collect your terms
    i expanded the  (x + y\sqrt2)^3 then do you substitute 7 + 5 \sqrt 2 into
    x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2
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  7. #7
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    x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}

    x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}

    So from here you can get the two equations I posted above.
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  8. #8
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    Quote Originally Posted by dkaksl View Post
    Not sure about the first one. But I think I got the second one for you.

    \frac{(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}

    \frac{(7-5\sqrt2)}{(7)^2-(5\sqrt2)^2}

    \frac{(7-5\sqrt2)}{49-50}

    \frac{(7-5\sqrt2)}{-1}

    -7+5\sqrt2

    Basically, you multiply and divide by the same thing. In the case of making the denominator (lower part of the fraction) simple, you multiply both nominator and denominator with (7-5\sqrt{2})

    Edit: Wait, you sure the book says \sqrt5 for the answer?
    yeh i made a typo its meant to be -7+5\sqrt2
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  9. #9
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    Quote Originally Posted by Stroodle View Post
    x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}

    x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}

    So from here you can get the two equations I posted above.
    well the book says the answer is 1+\sqrt2
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  10. #10
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by Paymemoney View Post
    well the book says the answer is 1+\sqrt2
    That's right. When you solve the two equations you get x=1 and y=1.
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  11. #11
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    ok but i still don't understand how you came up with the two equations from the equation x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2.

    i subsitutuded but it was unsuccessful.
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  12. #12
    Senior Member Stroodle's Avatar
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    When you arrange the equation into the form x^3+6xy^2+(2y^3+3x^2y)\sqrt{2}=7+5\sqrt{2}

    The coefficient of \sqrt{2} on the left hand side is 2y^3+3x^2y and the coefficient of the \sqrt{2} on the right hand side of the equation is 5

    So one equation to solve is 2y^3+3x^2y=5.

    Then the next equation is what is not the coefficient of \sqrt{2} on the left, equal to what is not the coeffient of \sqrt{2} on the right:

    x^3+6xy^2=7
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  13. #13
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    ok,
    firstly i understand how you got the two equations but the part that is confusing me is how did you get this equation: x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2 to

    this equation: x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}.
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  14. #14
    Senior Member Stroodle's Avatar
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    As posted above:

    x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}

    \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2} (from cubing both sides)

    \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}
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  15. #15
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Stroodle View Post
    For the first question:

    x+y\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}

    \Rightarrow (x+y\sqrt{2})^3=7+5\sqrt{2}

    \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}

    So from this you can now get two equations to solve simultaneously:

    x^3+6xy^2=7

    and 2y^3+3x^2y=5

    Hence x=1 and y=1

    \Rightarrow 1+\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}
    Quick question? How would you go about solving the 2 equations if it would not have been as easy as spotting that x=1 and y=1 works. Would you sub y=mx?
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