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Math Help - Finding the coefficients in a quadratic equation when you know the two roots

  1. #1
    Junior Member jonnygill's Avatar
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    Finding the coefficients in a quadratic equation when you know the two roots

    The problem is...

    Find a, b, c if...

    P(x)=ax^2+bx+c AND P(3+4i)=0 and P(3-4i)=0

    So, we know that 3+4i and 3-4i are two roots (imaginary).

    The way I approached this problem is I looked at the quadratic formula and I set a to a value such that the denominator equals 1.

    So, I let a=.5

    The formula becomes -b\pm\sqrt{b^2-2c}

    I figured b must equal -3 so that -b=3.

    I also figured that c must equal 12.5 so that the radical becomes \sqrt{-16}

    So i got...

    a=.5
    b=-3
    c=12.5

    The answer in the back of the book gives an answer of...

    a=1
    b=-6
    c=25

    Basically, they doubled all the coefficients. Are both answers acceptable? If only the books answer is acceptable, how could I solve this problem and arrive at the answer the book provided?

    Thanks.
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  2. #2
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    If \displaystyle 3 + 4i and \displaystyle 3 - 4i are roots of \displaystyle P(x) = ax^2 + bx + c, then

    \displaystyle a(3 + 4i)^2 + b(3 + 4i) + c = 0 and \displaystyle a(3 - 4i)^2 + b(3 + 4i) + c = 0.


    Simplifying equation 1: \displaystyle a(-7 + 24i) + 3b + 4bi + c = 0

    \displaystyle -7a + 24ai + 3b + 4bi + c = 0

    \displaystyle -7a+3b+c + (24a + 4b)i = 0

    Therefore \displaystyle -7a + 3b + c = 0 and \displaystyle 24a + 4b = 0 when you equate real and imaginary parts.


    Now do the same with Equation 2, and solve the resulting system for \displaystyle a,b,c.
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  3. #3
    Super Member TheChaz's Avatar
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    Wow. That's intense!
    x = 3 + 4i
    x - 3 = 4i
    Square both sides...
    x^2 - 6x + 9 = -16
    x^2 - 6x + 25.
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  4. #4
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    Is that exactly how the problem is worded? Because
    a\bigl(x-(3+4i)\bigr)\bigl(x-(3-4i)\bigr)
    will have the same zeros for any value of a.

    Your solution works, as does the book's.
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