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Thread: help with equation

  1. #1
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    help with equation

    {y^4+x^2-4=0
    {2y^2-x^2+1=0

    i need to find how many solutions can be, and to mark them on a graph.

    and then to know the zone of the shape that created as a result from the
    mark.

    hope u can understand my question.
    and yes my English is not so good.
    thank you all.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ZOOZ View Post
    {y^4+x^2-4=0
    {2y^2-x^2+1=0

    i need to find how many solutions can be, and to mark them on a graph.

    and then to know the zone of the shape that created as a result from the
    mark.

    hope u can understand my question.
    and yes my English is not so good.
    thank you all.
    From the first equation, you see that $\displaystyle \displaystyle x^2 = 4 - y^4$. Replace $\displaystyle \displaystyle x^2$ in the second equation with this expression. you get

    $\displaystyle \displaystyle 2y^2 - (4 - y^4) + 1 = 0$

    Now solve that equation for $\displaystyle \displaystyle y$. Then plug each of those y-values into the first equation to get the corresponding x-values.

    Can you finish? you should have 4 points as your answer.
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  3. #3
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    well thanks for ur hint but i don't know how to keep form this point 2y^2-3+y^4

    what i am doing with the y^4?

    hope u can help me
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ZOOZ View Post
    well thanks for ur hint but i don't know how to keep form this point 2y^2-3+y^4

    what i am doing with the y^4?

    hope u can help me
    Think of it as a quadratic equation:

    $\displaystyle \displaystyle y^4 + 2y^2 - 3 = 0$

    Is the same as saying

    $\displaystyle \displaystyle (y^2)^2 + 2(y^2) - 3 = 0$

    which is a quadratic equation in $\displaystyle \displaystyle y^2$. If you have trouble seeing this, let $\displaystyle \displaystyle a = y^2$, then the equation becomes

    $\displaystyle \displaystyle a^2 + 2a - 3 = 0$

    Solve that for $\displaystyle \displaystyle a$, then replace the $\displaystyle \displaystyle a$ with $\displaystyle \displaystyle y^2$ and solve for $\displaystyle \displaystyle y$
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  5. #5
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    hi i did as u told me and i got X1=1 X2=-3

    but i know i need to get 4 point so what i need to do?

    thanks for ur help.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ZOOZ View Post
    hi i did as u told me and i got X1=1 X2=-3

    but i know i need to get 4 point so what i need to do?

    thanks for ur help.
    what? what are x1 and x2??

    i think you mean you got $\displaystyle \displaystyle y^2 = 1$ or $\displaystyle \displaystyle y^2 = -3$. Clearly the last one is not possible for real numbers, so just take the first one. Then that means $\displaystyle \displaystyle y = \pm 1$. Plug each of these values into the first equation and solve for $\displaystyle \displaystyle x$. you will get 2 x-values each, giving you a total of four points.
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  7. #7
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    well ok if i take the value 1 from the y=+-1
    and put him in the first equation is will give me X^2=3

    let me know if u mean to take it and put it instead of y^4 ?

    appreciate ur patience
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ZOOZ View Post
    well ok if i take the value 1 from the y=+-1
    and put him in the first equation is will give me X^2=3

    let me know if u mean to take it and put it instead of y^4 ?

    appreciate ur patience
    if $\displaystyle \displaystyle x^2 = 3$, then what is $\displaystyle \displaystyle x$?
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  9. #9
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    hi thanks so much i success to solve it
    so really thank u.
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