Results 1 to 12 of 12

Math Help - using substitution to find coordinates of two quadratic curves, problems

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    16

    using substitution to find coordinates of two quadratic curves, problems

    i cannot understand where i am going wrong, the question is to find where the two curves
    meet:
    (squared=)

    x+y=25 (1)
    y=x-5 (2)

    heres how i worked it out:
    (1) y=25-x
    y=5-x
    subst y into 2
    5-x=x-5
    x+x-10=0
    this does not factorise! help! i am mathematically challenegd and i work hard to try abd get good grades and things like this frustrate me, i can do all the other simluatneous eqns except this type
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by chartsy View Post
    i cannot understand where i am going wrong, the question is to find where the two curves
    meet:

    next time, use the caret symbol (^) to show an exponent , not an emoticon.

    x^2 + y^2 = 25 (1)
    y=x^2 - 5 (2)
    from equation (2) ...

    x^2 = y + 5

    substitute y+5 for x^2 in equation (1) ...

    (y+5) + y^2 = 25

    y^2 + y - 20 = 0

    factor, solve for y, then determine the value of x.

    also, fyi ... if y^2 = 25 - x^2 , then y does not equal 5 - x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by chartsy View Post
    i cannot understand where i am going wrong, the question is to find where the two curves
    meet:
    (squared=)

    x+y=25 (1)
    y=x-5 (2)

    heres how i worked it out:
    (1) y=25-x

    y=5-x this is your error

    subst y into 2
    5-x=x-5
    x+x-10=0
    this does not factorise! help! i am mathematically challenegd and i work hard to try abd get good grades and things like this frustrate me, i can do all the other simluatneous eqns except this type
    x^2+y^2=25=5^2

    y=x^2-5\Rightarrow\ x^2=y+5

    Now you can make the substitution!

    y^2+y+5=25

    Now you solve the quadratic..

    Remember!! y^2=25-x^2\Rightarrow\ y=\pm\sqrt{25-x^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    16
    Quote Originally Posted by skeeter View Post
    from equation (2) ...

    x^2 = y + 5

    substitute y+5 for x^2 in equation (1) ...

    (y+5) + y^2 = 25

    y^2 + y - 20 = 0

    factor, solve for y, then determine the value of x.

    also, fyi ... if y^2 = 25 - x^2 , then y does not equal 5 - x
    ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by chartsy View Post
    ?
    what is your problem ? ... can you not factor y^2 + y - 20 = 0 ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2010
    Posts
    16
    o, wow i've got to read up
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    16
    Quote Originally Posted by skeeter View Post
    what is you problem ? ... can you not factor y^2 + y - 20 = 0 ?
    i can do that easily it was just what you said the y^2=25-x^2 isn't y=5-x
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    y^2=25-x^2=(5+x)(5-x)

    Only the square root of (5+x)(5+x) is 5+x.

    For example

    5^2=3^2+4^2 but 5\ \ne\ 3+4

    You have to take the square root of the entire RHS, not the square roots of each part of it...

    5=\sqrt{3^2+4^2}

    so if y^2=25-x^2\Rightarrow\ y=\pm\sqrt{25-x^2}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by chartsy View Post
    i can do that easily it was just what you said the y^2=25-x^2 isn't y=5-x
    what if x = 4 , what would you get for y ?

    do both equations give the same result?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Oct 2010
    Posts
    16
    so if i have to solve two quadratic curves simultaneously i should try and find the value of x/y with the equation which only has one value squared
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Oct 2010
    Posts
    16
    Quote Originally Posted by skeeter View Post
    what if x = 4 , what would you get for y ?

    do both equations give the same result?
    ok, so i will have to take the square root of the whole of the other sidem rather than the individual values
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by chartsy View Post
    so if i have to solve two quadratic curves simultaneously i should try and find the value of x/y with the equation which only has one value squared
    No, you only need to solve one quadratic.

    When you have found y, you can use it's value(s) to find x.

    x^2+y^2=25

    y=x^2-5\Rightarrow\ x^2=y+5\Rightarrow\ y^2+y+5=25

    You need to solve this for y.
    Then you can find x using the y solution(s).
    Last edited by Archie Meade; November 8th 2010 at 02:59 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral...substitution coordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 18th 2010, 06:15 AM
  2. coordinates of quadratic function
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 13th 2010, 04:30 AM
  3. Quadratic Equation from Coordinates?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 13th 2010, 02:07 AM
  4. Replies: 2
    Last Post: November 14th 2009, 04:40 PM
  5. U-Substitution and Area Between Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 11th 2008, 12:33 PM

Search Tags


/mathhelpforum @mathhelpforum