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Math Help - Finding the Value of x^2-y^2

  1. #1
    Junior Member lanierms's Avatar
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    Finding the Value of x^2-y^2

    I have this problem here which I can't seem to solve (why is my intro always like this):

    \sqrt{(x^2+y^2-20)^2}+[x^2+y^2+xy-18]=0

    The expression inside the [ ] means the absolute value. I can't seem to express absolute values, so.

    When real number x and y satisfies the above equation, what's the value of
    x^2-y^2?
    (x < y < 0)

    Can anyone help me out here?
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  2. #2
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    Quote Originally Posted by lanierms View Post
    I have this problem here which I can't seem to solve (why is my intro always like this):

    \sqrt{(x^2+y^2-20)^2}+[x^2+y^2+xy-18]=0

    The expression inside the [ ] means the absolute value. I can't seem to express absolute values, so.

    When real number x and y satisfies the above equation, what's the value of
    x^2-y^2?
    (x < y < 0)

    Can anyone help me out here?
    1. Both summands are positive. Thus this equation can only be true if both summands equal zero.

    2. From x^2+y^2-20=0 follows x^2+y^2=20. Plug in this value into the 2nd summand:

    x^2+y^2+xy-18=0~\implies~20+xy-18=0~\implies~y=-\frac2x

    3. The first equation becomes:

    x^2+\frac4{x^2}-20=0~\implies~x^4-20x^2+4=0

    Use the substitution z = x^2~\implies~|x|=\sqrt{z}

    Solve for z:

    z^2-20z+4=0~\implies~z=10\pm\sqrt{96}

    4. Therefore |x|=\sqrt{10\pm\sqrt{96}}. Plug in the appropriate values into the equation at #2.
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  3. #3
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    \sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-15| = 0

    \sqrt{(x^2+y^2-20)^2} = - (|x^2+y^2+xy-15|)

    (x^2+y^2-20)^2 = (|x^2+y^2+xy-15|)^2

    (x^2+y^2-20)^2 -(x^2+y^2+xy-15)^2= 0

    [(x^2+y^2-20) + (x^2+y^2+xy-15)][(x^2+y^2-20) - (x^2+y^2+xy-15)] = 0

    [2x^2 + 2y^2 + xy -35][-20 -xy +15] = 0

    [-20 -xy +15] = 0 Or xy = -5

    [2x^2 + 2y^2 + xy -35] = 0. Put xy = -5

    [2x^2 + 2y^2 -5 -35] = 0 Or

    <b>x^2 + y^2 = 20</b>

    (x^2 - y^2)^2 = (x^2+y^2)^2 - 4x^2y^2

    Substitute the values and solve for (x^2 - y^2)
    Last edited by sa-ri-ga-ma; June 9th 2010 at 03:24 AM.
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  4. #4
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    \sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-15| = 0

    \sqrt{(x^2+y^2-20)^2} = - (|x^2+y^2+xy-15|)

    (x^2+y^2-20)^2 = (|x^2+y^2+xy-15|)^2

    (x^2+y^2-20)^2 -(x^2+y^2+xy-15)^2= 0

    [(x^2+y^2-20) + (x^2+y^2+xy-15)][(x^2+y^2-20) - (x^2+y^2+xy-15)] = 0

    [2x^2 + 2y^2 + xy -35][-20 -xy +15] = 0

    [-20 -xy +15] = 0 Or xy = -5

    [2x^2 + 2y^2 + xy -35]= 0. Put xy = -5

    [2x^2 + 2y^2 -5 -35]= 0 Or

    x^2 + y^2 = 20

    (x^2 - y^2)^2 = (x^2+y^2)^2 - 4x^2y^2

    Substitute the values and solve for (x^2 - y^2)
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  5. #5
    Junior Member lanierms's Avatar
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    Thank you everyone for helping out my problem here.
    earboth's solution was the easiest to understand.
    thank you.
    oh yeah, how do you express the absolute value using Latex?
    Sorry for putting this topic under Algebra. I didn't know this was Pre-Calculus.
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by lanierms View Post
    oh yeah, how do you express the absolute value using Latex?
    On my keyboard, the key is right above "enter" on the same key as backslash (\); so it's shift-backslash.
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  7. #7
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    Quote Originally Posted by lanierms View Post
    Thank you everyone for helping out my problem here.
    earboth's solution was the easiest to understand.
    thank you.
    ... but didn't answer your question
    Quote Originally Posted by lanierms View Post
    ...

    When real number x and y satisfies the above equation, what's the value of
    \bold{\color{blue}x^2-y^2}
    ?
    (x < y < 0)

    Can anyone help me out here?
    Quote Originally Posted by sa-ri-ga-ma View Post
    \sqrt{(x^2+y^2-20)^2}+ |x^2+y^2+xy-\bold{\color{red}15}| = 0

    ...
    Unfortunately sa-ri-ga-ma made a typo.



    If you use
    x^2+y^2=20~\wedge~xy=-2

    then you'll get:

    x^2+y^2=\sqrt{20^2-4 \cdot 4}= 8\cdot \sqrt{6}
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  8. #8
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    Quote Originally Posted by earboth View Post
    Unfortunately sa-ri-ga-ma made a typo.



    If you use
    x^2+y^2=20~\wedge~xy=-2

    then you'll get:

    x^2+y^2=\sqrt{20^2-4 \cdot 4}= 8\cdot \sqrt{6}
    Thank you earboth. In that case, the equation



    becomes
    (2x^2 + 2y^2 - 38)(20 - xy + 18) = 0

    So

    x^2 + y^2 = 19; and ;xy = -2

    x^2 - y^2 = \sqrt{19^2 - 4\cdot4)}
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  9. #9
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Thank you earboth. In that case, the equation



    becomes
    (2x^2 + 2y^2 - 38)(20 - xy + 18) = 0

    So

    x^2 + y^2 = 19; and ;xy = -2

    x^2 - y^2 = \sqrt{19^2 - 4\cdot4)}
    Next step of the endless story:

    In that case the equation:

    \left((8x^2+y^2-20)+(x^2+y^2+xy-18)\right) \left((8x^2+y^2-20) -(x^2+y^2+xy-18)\right) =0

    becomes:

    (2x^2+2y^2\bold{+xy}-38)(20+xy-18)=0

    If the the 2nd bracket equals zero then xy = -2.

    Therefore the 1st bracket is:

    (2x^2+2y^2-2-38)=0~\implies~x^2+y^2=20

    etc. andsoon ...
    Last edited by earboth; June 20th 2010 at 09:04 AM.
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  10. #10
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    Quote Originally Posted by earboth View Post
    Next step of the endless story:

    In that case the equation:

    \left((8x^2+y^2-20)+(x^2+y^2+xy-18)\right)\left((8x^2+y^2-20) -(x^2+y^2+xy-18)\right) =0

    becomes:

    (2x^2+2y^2\bold{\color{blue}+xy}-38)(20+xy-18)=0

    If the the 2nd bracket equals zero then xy = -2.

    Therefore the 1st bracket is:

    (2x^2+2y^2-2-38)=0~\implies~x^2+y^2=20

    etc. andsoon ...
    Oooooo No! Thanks.
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