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Thread: 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
    $\displaystyle \alpha = \frac{-b+\sqrt{b^2-4ac}}{2a}$

    $\displaystyle \beta = \frac{-b-\sqrt{b^2-4ac}}{2a}$
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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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    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 $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the quadratic equation $\displaystyle ax^2+bx+c=0$, then:

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

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

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

    Part (ii)

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

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

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

    Part (iii)

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

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

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

    $\displaystyle = 11\alpha+10$
    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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