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    Basically, I need to find the stationary points of the equation x^2+y^2+xy=3

    This is how far I've gotten,

    x^2+y^2+xy=3

    Differentiating with respect to x

    2x+2y.\dfrac{dy}{dx}+x.\dfrac{dy  }{dx}+y=0

    Then, bringing the \dfrac{dy}{dx} together, I get,

    \dfrac{dy}{dx}(2y+x)=-2x-y

    \dfrac{dy}{dx}=\dfrac{-2x-y}{2y+x}

    At stationary points, \dfrac{dy}{dx}=0, so,

    \dfrac{-2x-y}{2y+x}=0

    -2x-y=0

    y=-2x

    I don't see where I can go from here and how I can find the stationary points. I need help please?

    Thanks.
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    (Original post by RamocitoMorales)
    x^2+y^2+xy=3

    y=-2x
    You have two equations there, what can you do with them?
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    Where (on that curve) is y equal to -2x?
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    (Original post by LetoKynes)
    You have two equations there, what can you do with them?
    Simultaneously solve it?
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    (Original post by LetoKynes)
    You have two equations there, what can you do with them?
    It still doesn't work, this is what happens.

    x^2+y^2+xy=3

    y=-2x

    Substituting in the y value into the first equation, I get,

     x^2+(-2x)^2+x(-2x)=3

    x^2+4x^2-4x^2=3

    x^2=3

    x=\sqrt{3}

    That's not the right answer, the right answer should be (1,-2), (-1, 2).
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    Since when has x(-2x) equalled -4x^2?
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    (Original post by RamocitoMorales)
    It still doesn't work, this is what happens.

    x^2+y^2+xy=3

    y=-2x

    Substituting in the y value into the first equation, I get,

     x^2+(-2x)^2+x(-2x)=3

    x^2+4x^2-4x^2=3 Mistake on this line

    x^2=3

    x=\sqrt{3}

    That's not the right answer, the right answer should be (1,-2), (-1, 2).
    Do you get the right answer if you correct that?
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    (Original post by LetoKynes)
    Do you get the right answer if you correct that?
    I can be such a plum sometimes, sorry about that.

    So, continuing form the mistake, I get,

    x^2+4x^2-2x^2=3

    3x^2=3

    x^2=\dfrac{3}{3}

    x^2=1

    x=\sqrt{1}

    x=1 or x=-1

    Thank you, that is the correct answer for x. Now I will sub it back into the original equation to find the value for y.

    x^2+y^2+xy=3

    x=1 in this situation, so,

    1+y^2+y=3

    y^2+y=3-1

    y^2+y=2

    So now what do I do?

    Thanks.
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    Wiki Support Team
    Solve the quadratic equation...?
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    Yeap, just solve the quadratic equation.
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    (Original post by generalebriety)
    Solve the quadratic equation...?
    So basically,

    y^2+y=2

    Which can be simplified to,

    y^2+y-2=0

    Which can be factorised as,

    (y+2)(y-1)=0

    So y=1 or y=-2

    But it's still wrong, since on the answers it says that y should equal to -2 and 2, not -2 and 1.

    hmmmm

    I would give you reputation, but it's worthless.
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    You can use the x = -1 and you should get the other y value

    Although I'm not sure how you're supposed to know which value to use.. :p:
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    (Original post by LetoKynes)
    You can use the x = -1 and you should get the other y value

    Although I'm not sure how you're supposed to know which value to use.. :p:
    Ah of course, I'd forgotten all about the x=-1

    Now when x=-1 in the equation x^2+y^2+xy=3,

    (-1)^2+y^2+(-1)y=3

    1+y^2-y=3

    y^2-y+1-3=0

    y^2-y-2=0

    (y-2)(y+1)=0

    So, y=2, or y=-1

    This is a very confusing question.
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    (Original post by RamocitoMorales)
    Ah of course, I'd forgotten all about the x=-1

    Now when x=-1 in the equation x^2+y^2+xy=3,

    (-1)^2+y^2+(-1)y=3

    1+y^2-y=3

    y^2-y+1-3=0

    y^2-y-2=0

    (y-2)(y+1)=0

    So, y=2, or y=-1

    This is a very confusing question.
    It's not that confusing. If x = 1 then y = 1 or -2; if x = -1 then y = -1 or 2. Which values are valid (remembering that y = -2x)? They will then all satisfy the original equation, so you're done.
 
 
 
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