Results 1 to 5 of 5

Math Help - Rational numbers?

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    10

    Rational numbers?

    So yesterday, on math class, I solved a difficult task (a quadratic polynomial). I solved it correctly, but at one point teacher asked me something I couldn't explain. The task says: "The biggest value of the quadratic polynomial is 3/2 and one of his zeros is √3 - 1. Determine the polynomial if it's coefficients are rational numbers!". The first zero was √3 - 1, and I knew the other one had to be -√3 - 1 (I know it had to be it's conjugate). Teacher said that these are NOT complex numbers - so i cant conjugate anything. How can I prove that the rational coefficient [√3 - 1] is equal(?) to the rational coefficient [-√3 - 1]?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by IoNForce View Post
    So yesterday, on math class, I solved a difficult task (a quadratic polynomial). I solved it correctly, but at one point teacher asked me something I couldn't explain. The task says: "The biggest value of the quadratic polynomial is 3/2 and one of his zeros is √3 - 1. Determine the polynomial if it's coefficients are rational numbers!". The first zero was √3 - 1, and I knew the other one had to be -√3 - 1 (I know it had to be it's conjugate). Teacher said that these are NOT complex numbers - so i cant conjugate anything. How can I prove that the rational coefficient [√3 - 1] is equal(?) to the rational coefficient [-√3 - 1]?
    1. The quadratic equation y = ax^2+bx+c can be rearranged to y = a(x-x_v)^2+y_v where V(x_v, y_v) is the vertex of the corresponding parabola.

    2. For symmetry's reason you know that the mean of the zeros equals x_v. From \sqrt{3}-1 you can get two different possible zeros: \sqrt{3}+1 or -\sqrt{3}-1

    Calculating the mean value you'll get:

    \dfrac{(\sqrt{3}-1)+(\sqrt{3}+1)}2 = \sqrt{3} which is definitely no rational number;
    or
    \dfrac{(\sqrt{3}-1)+(-\sqrt{3}-1)}2 = -1 which is a valid number for x_v according to the text of your question.

    3. The maximum value occurs at the vertex. Therefore you know now:

    y = a(x+1)^2+\frac32

    4. Plug in the value of the known zero and solve for a:

    0=a(\sqrt{3}-1+1)^2+\frac32

    I've got a = -\frac12
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    10
    Quote Originally Posted by earboth View Post
    1. The quadratic equation y = ax^2+bx+c can be rearranged to y = a(x-x_v)^2+y_v where V(x_v, y_v) is the vertex of the corresponding parabola.

    2. For symmetry's reason you know that the mean of the zeros equals x_v. From \sqrt{3}-1 you can get two different possible zeros: \sqrt{3}+1 or -\sqrt{3}-1

    Calculating the mean value you'll get:

    \dfrac{(\sqrt{3}-1)+(\sqrt{3}+1)}2 = \sqrt{3} which is definitely no rational number;
    or
    \dfrac{(\sqrt{3}-1)+(-\sqrt{3}-1)}2 = -1 which is a valid number for x_v according to the text of your question.

    3. The maximum value occurs at the vertex. Therefore you know now:

    y = a(x+1)^2+\frac32

    4. Plug in the value of the known zero and solve for a:

    0=a(\sqrt{3}-1+1)^2+\frac32

    I've got a = -\frac12
    Yeah I've got a = -\frac12 too, but I have to prove the bold text. The reason why can I get the other two possible zeros from \sqrt{3}-1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, IoNForce!

    So yesterday, on math class, I solved a difficult task (a quadratic polynomial).
    I solved it correctly, but at one point teacher asked me something I couldn't explain.

    The task says: "The biggest value of the quadratic polynomial is 3/2
    and one of its zeros is √3 - 1.
    Determine the polynomial if it's coefficients are rational numbers."

    The first zero was √3 - 1, and I knew the other one had to be -√3 - 1.
    (I know it had to be its conjugate).
    Teacher said that these are NOT complex numbers - so i can't conjugate anything.
    Your teacher is wrong!

    You are right . . . Irrational nuimbers have conjugates, too.

    This is the way I approached it . . .


    The two roots are: . \text{-}1 + \sqrt{3}\,\text{ and }\,\text{-}1 - \sqrt{3}


    A quadratic polynomial with these two root has the form:

    . . f(x) \;=\;a\bigg[x - (-1 + \sqrt{3})\bigg]\,\bigg[x -(-1 - \sqrt{3})\bigg]

    . . . . . . =\;a\bigg[(x-1) - \sqrt{3}\bigg]\,\bigg[(x-1) + \sqrt{3})\bigg]

    . . . . . . =\;a\bigg[(x-1)^2 - (\sqrt{3})^2\bigg]

    . . . . . . =\;a(x^2 - 2x + 1 - 3) \;=\;a(x^2 - 2x - 2)


    We have: . f(x) \;=\;ax^2 - 2ax - 2a

    The maximum value (\frac{3}{2}) occurs at its vertex.

    The vertex is at: . x \:=\:\frac{-(\text{-}2a)}{2(a)} \:=\:1

    . . and: . f(1) \:=\:a(1^2) - 2a(1) - 2a \:=\:\text{-}3a


    Since the maximum vaue is \frac{3}{2}, we have: . \text{-}3a \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\text{-}\frac{1}{2}


    Therefore: . f(x) \;=\;\text{-}\frac{1}{2}(x^2 + 2x - 2) \;=\;\text{-}\frac{1}{2}x^2 - x + 1


    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1
    The two roots are -1+\sqrt{3}~\&~-1-\sqrt{3} .
    So the equation is x^2-[(-1+\sqrt{3})+(-1-\sqrt{3})]x+(-1+\sqrt{3})(-1-\sqrt{3})=0 .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof: All rational numbers are algebraic numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 5th 2010, 10:26 AM
  2. Rational numbers
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 15th 2009, 12:42 AM
  3. Replies: 8
    Last Post: September 15th 2008, 04:33 PM
  4. Rational Numbers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 14th 2007, 03:35 PM
  5. More rational numbers! Yay!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 8th 2007, 06:49 PM

Search Tags


/mathhelpforum @mathhelpforum