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Math Help - Roots of quadratic functions

  1. #1
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    Roots of quadratic functions

    If a and b are roots of the quadratic equation 3x^2-6x+2=0.
    (i) Find a+b
    (ii) Find ab

    Anyone can help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by acc100jt View Post
    If a and b are roots of the quadratic equation 3x^2-6x+2=0.
    (i) Find a+b
    (ii) Find ab

    Anyone can help?
    If a and b are the roots of 3x^2-6x+2, then:

    3(x-a)(x-b)=3x^2-6x+2

    Expand the left hand side and equate corfficients of x^2, and x, and the constant terms:

    3x^2 - 3(a+b)x + 3ab = 3x^2 - 6x + 2

    So (a+b)= 2, and ab=2/3.

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by acc100jt View Post
    If a and b are roots of the quadratic equation 3x^2-6x+2=0.
    (i) Find a+b
    (ii) Find ab

    Anyone can help?
    CaptainBlack's method is perfectly acceptable, but there is a general theorem for quadratics you can use to tackle this. (You can actually get theorems for this for higher order polynomials, but I forget the name of the method.)

    If you have two roots of a quadratic equation ax^2 + bx + c = 0, r_1~\text{and}~r_2, then

    r_1 + r_2 = -\frac{b}{a}
    and
    r_1 \cdot r_2 = \frac{c}{a}

    (You can easily get this result by noting that
    r_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} and doing the relevant sum or product of r_1~\text{and}~r_2.)

    In your particular case, we have a = 3, b = -6, and c = 2. Thus
    r_1 + r_2 = -\frac{-6}{3} = 2
    and
    r1 \cdot r_2 = \frac{2}{3}
    just as CaptainBlack derived.

    -Dan
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  4. #4
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    It is called "Viete's Theorem".
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