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Math Help - quadratic sinutanius equations

  1. #1
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    quadratic sinutanius equations

    hi,

    im not too sure how to solve this question:

    quadratic sinutanius equations-question.jpg

    can someone please tell me how to do this one!

    im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

    thanks!
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    hi,

    im not too sure how to solve this question:

    Click image for larger version. 

Name:	question.jpg 
Views:	44 
Size:	17.2 KB 
ID:	20670

    can someone please tell me how to do this one!

    im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

    thanks!
    Do you know how to substitute? What do your class notes/text reveal?
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  3. #3
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    no i don't know how to do it for this question because it has x^2 and y^2 in it!
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    no i don't know how to do it for this question because it has x^2 and y^2 in it!
    Solve for x (which is equal to 5 - y), substitute into the second equation (which is quadratic - I hope you know how to solve a quadratic equation). Once you figure out y, then you can get x from the first equation.
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  5. #5
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    simultaneous equations ...

    x+y = 5 ... this is a linear equation

    x^2+y^2 = 13 ... this is the equation of a circle

    using the linear equation, note that y = 5-x

    substitute (5-x) for  y in the circle equation ...

    x^2 + (5-x)^2 = 13<br />

    solve the quadratic equation for x , then you can determine y.

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  6. #6
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    so i would do this to find X:

    quadratic sinutanius equations-correct-3.jpg

    is this correct so far?
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  7. #7
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    and to get Y would i do:

    5-1.2= 3.8
    1.2^2= 1.44
    3.8^2= 14.44
    14.44-1.44=13

    Y= 3.8??
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  8. #8
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    Quote Originally Posted by andyboy179 View Post
    so i would do this to find X:

    Click image for larger version. 

Name:	is this correct 3.jpg 
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ID:	20671

    is this correct so far?
    Unfortunately no.

    1. According to the binomial formula

    (5-x)^2 = (5-x)(5-x) = 25-10x+x^2

    2. So your equation becomes

    2x^2-10x+12 = 0~\implies~x^2-5x+6=0~\implies~(x-2)(x-3)=0

    A product equals zero if one of the factors equals zero.
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  9. #9
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    Quote Originally Posted by andyboy179 View Post
    hi,

    im not too sure how to solve this question:

    Click image for larger version. 

Name:	question.jpg 
Views:	44 
Size:	17.2 KB 
ID:	20670

    can someone please tell me how to do this one!

    im okay at doing them with Y= at the start but haven't done one with x^2 and y^2 before.

    thanks!
    x+y=5

    x^2+y^2=13


    (x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)=x^2+xy+xy+y^2

    (x+y)^2=x^2+y^2+2xy=5^2=25

    (x+y)^2-\left(x^2+y^2\right)=2xy=25-13=12\Rightarrow\ xy=6

    So you could also solve

    x+y=5

    xy=6
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  10. #10
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    oh i see where i went wrong! i got -x^2 and it should be 2x^2. i can take it from here thanks!
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