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Math Help - Quadratic equation with imaginary roots

  1. #1
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    Quadratic equation with imaginary roots

    I have done part i of the attached question. Do I have to understand complex numbers for part ii?
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  2. #2
    Junior Member doomgaze's Avatar
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    Quote Originally Posted by Stuck Man View Post
    I have done part i of the attached question. Do I have to understand complex numbers for part ii?

    I don't see anything.

    Try to post the question here or take a picture of it?
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  3. #3
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    Attached Thumbnails Attached Thumbnails Quadratic equation with imaginary roots-scan-1.jpg  
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Stuck Man View Post
    \alpha = \frac{-b+\sqrt{b^2-4ac}}{2a}

    \beta = \frac{-b-\sqrt{b^2-4ac}}{2a}
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  5. #5
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    b^2-4ac is -23. I don't know how to get a square root of a negative so thats why I'm asking if this requires a knowledge of complex numbers.
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Stuck Man View Post
    b^2-4ac is -23. I don't know how to get a square root of a negative so thats why I'm asking if this requires a knowledge of complex numbers.
    Ah right, my mistake.

    Yes, complex number knowledge is necessary IMO. Perhaps someone better than me can explain a way around it?
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  7. #7
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    I've solved ii now. It takes quite a lot of writing. Square roots of -23 get squared or cancel out near the end.
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  8. #8
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    Hello Stuck Man

    It looks as if you don't know about this important result:

    • If \alpha and \beta are the roots of the quadratic equation ax^2+bx+c=0, then:

    The sum of its roots, \alpha + \beta = -\frac{b}{a}
    and
    The product of its roots, \alpha\beta = \frac{c}{a}
    (This is very easy to prove: expand (x-\alpha)(x-\beta) and compare it to x^2 + \frac{b}{a}x + \frac{c}{a}.)

    Well then, if \alpha and \beta are the roots of 3x^2 +x + 2 = 0, we get:
    \alpha + \beta = -\frac{1}{3}
    and
    \alpha\beta = \frac{2}{3}
    So for part (i):
    \frac{1}{\alpha^2}+\frac{1}{\beta^2} =\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}
    = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}
    Now substitute in the values of \alpha + \beta and \alpha\beta, simplify and the job's done. (I get the answer -\frac{11}{4}. Do you agree?)

    ****************

    Part (ii)

    We know the value of \frac{1}{\alpha^2}+\frac{1}{\beta^2} and clearly \frac{1}{\alpha^2}\times\frac{1}{\beta^2}= \frac{1}{(\alpha\beta)^2}= \frac94. So we can use the result in reverse to write down the new equation, knowing the sum and product of its roots; i.e.
    x^2 - \text{(sum of roots)}x+ \text{(product of roots)}=0
    which (if my arithmetic is correct) is:
    x^2 +\frac{11}{4}x+\frac94=0

    i.e. 4x^2+11x+9=0

    ****************

    Part (iii)

    \alpha is a root of 3x^2+x+2=0. So:
    3\alpha^2+\alpha+2=0

    \Rightarrow 3\alpha^2 = -\alpha-2
    (1)

    \Rightarrow 9\alpha^4 =(-\alpha-2)^2
    =\alpha^2 +4\alpha+4
    \Rightarrow 27\alpha^4 = 3\alpha^2 +12\alpha+12
    =-\alpha - 2 +12 \alpha +12 from (1)

    = 11\alpha+10
    Grandad
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  9. #9
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    I did part i the same. For parts ii and iii I used the quadratic formula to get a value for alpha. I got the answers but it was far more work.


    Thanks.
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