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Thread: 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 $\displaystyle x+y\sqrt2$ with x and y rational numbers:
    1)$\displaystyle (7+5\sqrt2)^{\frac{1}{3}}$

    Express the following numbers in the form $\displaystyle x+y\sqrt{n}$ where x and y are rational numbers and n is an integer:
    2)$\displaystyle \frac{1}{7+5\sqrt2}$
    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 $\displaystyle -7+5\sqrt5$:
    $\displaystyle =\frac{1(7+5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

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

    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.

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

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

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

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

    $\displaystyle -7+5\sqrt2$

    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 $\displaystyle (7-5\sqrt{2})$

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

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

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

    Expand the left hand side and collect your terms

    __________________________________________________

    Express the following numbers in the form $\displaystyle x+y\sqrt{n}$ 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)

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

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

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

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

    $\displaystyle = -7 + 5\sqrt2 $
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  4. #4
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    thanks for the solutions.
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  5. #5
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    For the first question:

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

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

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

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

    $\displaystyle x^3+6xy^2=7$

    and $\displaystyle 2y^3+3x^2y=5$

    Hence $\displaystyle x=1$ and $\displaystyle y=1$

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

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

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

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

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

    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.

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

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

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

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

    $\displaystyle -7+5\sqrt2$

    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 $\displaystyle (7-5\sqrt{2})$

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

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

    So from here you can get the two equations I posted above.
    well the book says the answer is $\displaystyle 1+\sqrt2$
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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 $\displaystyle 1+\sqrt2$
    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 $\displaystyle x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$.

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

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

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

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

    $\displaystyle x^3+6xy^2=7$
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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:$\displaystyle x^3+3x^2y\sqrt2+3xy^2\sqrt2+y^3\sqrt2$ to

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

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

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

    $\displaystyle \Rightarrow x^3+2\sqrt{2}y^3+3\sqrt{2}x^2y+6xy^2=7+5\sqrt{2}$
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  15. #15
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    Quote Originally Posted by Stroodle View Post
    For the first question:

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

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

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

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

    $\displaystyle x^3+6xy^2=7$

    and $\displaystyle 2y^3+3x^2y=5$

    Hence $\displaystyle x=1$ and $\displaystyle y=1$

    $\displaystyle \Rightarrow 1+\sqrt{2}=(7+5\sqrt{2})^{\frac{1}{3}}$
    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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