Results 1 to 6 of 6

Math Help - Fixed Real Numbers...Part 1

  1. #1
    Banned
    Joined
    Dec 2009
    From
    NYC
    Posts
    84

    Fixed Real Numbers...Part 1

    Suppose a, b, and c are fixed real numbers such that
    b^2 - 4ac ≥ 0. Let r and s be the solutions of
    ax^2 + bx + x = 0.

    (a) Use the quadratic formula to show that r + s = -b/a and
    rs = c/a.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by sologuitar View Post
    Suppose a, b, and c are fixed real numbers such that
    b^2 - 4ac ≥ 0. Let r and s be the solutions of
    ax^2 + bx + x = 0.

    (a) Use the quadratic formula to show that r + s = -b/a and
    rs = c/a.
    ax^2+bx+c = a(x-R)(x-S)

    <br />
\Rightarrow ax^2+bx+c = ax^2-a(R+S)x+aRS

    Comparing coefficients gives

    b = -a(R+S) \rightarrow R+S = \frac{-b}{a}

    and

    c = aRS \rightarrow RS = \frac{c}{a}

    Hope this helps!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by sologuitar View Post
    Suppose a, b, and c are fixed real numbers such that
    b^2 - 4ac ≥ 0. Let r and s be the solutions of
    ax^2 + bx + x = 0.

    (a) Use the quadratic formula to show that r + s = -b/a and
    rs = c/a.

    ax^2 + bx + c

    a\left( x^2 + \frac{b}{a} \cdot x + \frac{c}{a} \right)

    r,s are roots of the equation so we can write the equation like this

    (x-r)(x-s)= x^2 + \frac{b}{a} \cdot x + \frac{c}{a}

    x^2 +x(-r-s) + rs = x^2 + \frac{b}{a} \cdot x + \frac{c}{a}

    -r-s = \frac{b}{a} \Rightarrow r+s = \frac{-b}{a}

    rs = \frac{c}{a}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,396
    Thanks
    1846
    Oh, this was a! I did the problems in the wrong order!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Dec 2009
    From
    NYC
    Posts
    84

    ok...

    Quote Originally Posted by Amer View Post
    ax^2 + bx + c

    a\left( x^2 + \frac{b}{a} \cdot x + \frac{c}{a} \right)

    r,s are roots of the equation so we can write the equation like this

    (x-r)(x-s)= x^2 + \frac{b}{a} \cdot x + \frac{c}{a}

    x^2 +x(-r-s) + rs = x^2 + \frac{b}{a} \cdot x + \frac{c}{a}

    -r-s = \frac{b}{a} \Rightarrow r+s = \frac{-b}{a}

    rs = \frac{c}{a}
    I would have not been able to work my way through this question.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Dec 2009
    From
    NYC
    Posts
    84

    thanks...

    Quote Originally Posted by BabyMilo View Post
    ax^2+bx+c = a(x-R)(x-S)

    <br />
\Rightarrow ax^2+bx+c = ax^2-a(R+S)x+aRS

    Comparing coefficients gives

    b = -a(R+S) \rightarrow R+S = \frac{-b}{a}

    and

    c = aRS \rightarrow RS = \frac{c}{a}

    Hope this helps!
    This question is interestingly hard.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. real part of a function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 24th 2010, 11:44 PM
  2. Finding fixed numbers
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 8th 2010, 11:47 PM
  3. Finding the Real/Imaginary Part of f(z), g(z)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2010, 02:56 PM
  4. Fixed Real Numbers...Part 3
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 8th 2009, 08:02 AM
  5. Fixed Real Numbers...Part 2
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 8th 2009, 08:00 AM

Search Tags


/mathhelpforum @mathhelpforum