Results 1 to 13 of 13

Math Help - parabola problems, and other questions

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    11

    parabola problems, and other questions

    Hello, I have a few questions that are making my head hurt.

    Question 1:
    if the parabola y -2=a(x-3)^2 goes through the point (2,0), what is the value of a?

    Question 2:
    Given x^3-4x^2+2x+1=0
    a:How many possible positive roots are there?
    b:How many possible negative roots are there?
    c:What are the possible negative roots?
    d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

    Thanks
    Last edited by koumori; January 29th 2010 at 02:48 PM. Reason: fixing bad spelling.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    180

    First one..

    Given y-2 = a(x-3)^2 /devide both sides by (x-3)^2

    then a = (y-2)/(x-3)^2 you know that when x = 2; y = 0 so just plug those in.. and there you have it...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by koumori View Post
    Question 1:
    if the parabola y -2=a(x-3)^2 goes through the4 point (2,0), what is the value of a?


    Thanks
    0 -2=a(2-3)^2

    now solve for a . You may need to use complex numbers.


    Quote Originally Posted by koumori View Post

    Question 2:
    Given x^3-4x^2+2x+1=0
    a:How many possible positive roots are there?
    b:How many possible negative roots are there?
    c:What are the possible negative roots?
    d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

    Thanks

    This one is a little harder, what have you tried?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    180

    how?

    Just curious, how could a complex number be involved?

    I just get a = -2?

    what am I missing?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by koumori View Post
    Hello, I have a few questions that are making my head hurt.

    Question 1:
    if the parabola y -2=a(x-3)^2 goes through the point (2,0), what is the value of a?

    Question 2:
    Given x^3-4x^2+2x+1=0
    a:How many possible positive roots are there?
    b:How many possible negative roots are there?
    c:What are the possible negative roots?
    d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

    Thanks
    2) you know something times something = 1
    so you try 1 for example,
    1^3-4(1)^2+2(1)+1=0

    so (x-1) is one of the root.

    divide the equation by (x-1)
    (x-1)(x^2-3x-1)=0

    use the quadratic equation. to solve for the other two roots.

    2a) 2
    b)1
    e) \frac{3+/-\sqrt{(-3)^2-4*1*-1)}}{2}
    =\frac{3+/-\sqrt{13}}{2}

    hope this helps.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Quote Originally Posted by pickslides View Post
    Quote Originally Posted by koumori View Post
    Question 2:
    Given x^3-4x^2+2x+1=0
    a:How many possible positive roots are there?
    b:How many possible negative roots are there?
    c:What are the possible negative roots?
    d:Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    e:Find the irrational roots of the equation.(Hint:Use the quadratic formula to solve the depressed equation.)

    This one is a little harder, what have you tried?
    Well...I'm pretty sure I know the answers for a,b, and c: (a:1)(b:1)(c:-\frac{1}{1} , \frac{1}{1}.
    For "d" I've used synthetic substitution and I think that the answer is -1, that last one is the one that makes my head hurt.
    If I'm not much mistaken, the depressed equation is: [tex]x^2+3x+1=0[\math]
    when I use the quadratic equation on it, I get a gibberish answer that is rational:-2.168 approximately.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by koumori View Post
    Well...I'm pretty sure I know the answers for a,b, and c: (a:1)(b:1)(c:-\frac{1}{1} , \frac{1}{1}.
    For "d" I've used synthetic substitution and I think that the answer is -1, that last one is the one that makes my head hurt.
    If I'm not much mistaken, the depressed equation is: [tex]x^2+3x+1=0[\math]
    when I use the quadratic equation on it, I get a gibberish answer that is rational:-2.168 approximately.
    x = 1 is clearly a solution. Therefore x - 1 is a factor of x^3 - 4x^2 + 2x + 1. Therefore the quadratic factor is x^2 - 3x - 1.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2010
    Posts
    11
    ok, I think I really went wrong somewhere!
    When I do synthetic division on the equation I find that -1 gives a remainder of 0....
    I remember that if there is no remainder left after doing synthetic division then the divisor is a factor of the equation.

    Now I am confused.....where did I mess up?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by koumori View Post
    ok, I think I really went wrong somewhere!
    When I do synthetic division on the equation I find that -1 gives a remainder of 0....
    I remember that if there is no remainder left after doing synthetic division then the divisor is a factor of the equation.

    Now I am confused.....where did I mess up?
    Since you have not posted your working it's impossible to know where you messed up (and yes, you have messed up somewhere because (-1)^3 - 4(-1)^2 + 2(-1) + 1 does not equal zero).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Ok....lesson learned!
    1 is the number to use, so that means that the depressed equation is:
    x^2 - 3x - 1 then, the result, when the quadratic equation is applied is 3+i or 3-i?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Dec 2009
    From
    U.S
    Posts
    60
    Quote Originally Posted by pickslides View Post
    0 -2=a(2-3)^2

    now solve for a . You may need to use complex numbers.





    This one is a little harder, what have you tried?
    Complex numbers are unnecessary to solve the equation you posted. -1 squared is 1. Therefore you have -2 = a*1; thus a = \frac {-1}{2}.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Feb 2010
    Posts
    17
    Try using Descartes' Rule of Signs
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Quote Originally Posted by mattio View Post
    Try using Descartes' Rule of Signs
    Thanks! I did, I got an answer of 1 possible positive root and one possible negative root.
    Is that correct?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with parabola questions for project
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 2nd 2009, 11:55 PM
  2. parabola problems
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 1st 2008, 02:53 PM
  3. Parabola questions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 19th 2007, 04:18 PM
  4. parabola questions ( please help!)
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 16th 2007, 11:33 PM
  5. Parabola problems
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 14th 2007, 07:00 PM

Search Tags


/mathhelpforum @mathhelpforum