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Thread: Values for alpha and beta.

  1. #1
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    Values for alpha and beta.

    Hello

    Let $\displaystyle \alpha,\beta$ be the two solutions of the quadratic equation

    $\displaystyle 3x^2+6x+7=0.$

    Then the value of $\displaystyle (2\alpha-\beta)(2\beta-\alpha)$ is ?


    How can I get the values for $\displaystyle \alpha$ and $\displaystyle \beta$? After trying with the quadratic ecuation, I was always getting a negative value on the $\displaystyle sqrt{}$ section.

    Thanks.
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  2. #2
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    You got negative values in the square root because this quadratic has two complex solutions. Continue on your way.

    Checck the discriminat, $\displaystyle b^{2}-4ac$. If it's negative, there are no real roots.

    $\displaystyle (6)^{2}-4(3)(7)=-48$.........yep, negative discriminant, so only complex solutions.
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  3. #3
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    Hello, Patrick_John!

    There is a back-door approach to this problem
    . . if you know some polynomial theory.


    Let $\displaystyle \alpha,\beta$ be the two solutions of the quadratic equation: .$\displaystyle 3x^2+6x+7\:=\:0$

    Find the value of: .$\displaystyle (2\alpha-\beta)(2\beta-\alpha)$
    We want the value of: .$\displaystyle V \;=\;(2\alpha-\beta)(2\beta-\alpha) \;=\;5\alpha\beta - 2(\alpha^2 + \beta^2)$ .[A]


    The equation is: .$\displaystyle x^2 + 2x + \frac{7}{3}\:=\:0$

    Then: .$\displaystyle \begin{array}{cccc}\alpha + \beta & = & -2 & [1]\\ \alpha\beta & = & \frac{7}{3}& [2]\end{array}$

    $\displaystyle \text{Square [1]: }\;(\alpha +\beta)^2\:=\:(-2)^2\quad\Rightarrow\quad\alpha^2 + 2\!\!\underbrace{(\alpha\beta)}_{\text{this is }\frac{7}{3}} + \beta^2 \:=\:4$
    . . So we have: .$\displaystyle \alpha^2 + 2\left(\frac{7}{3}\right) + \beta^2\:=\:4\quad\Rightarrow\quad\alpha^2 + \beta^2\:=\:-\frac{2}{3}\;\;\;[3]$

    Substitute [2] and [3] into equation [A]: .$\displaystyle V \;=\;5\left(\frac{7}{3}\right) - 2\left(-\frac{2}{3}\right) \;=\;\boxed{13}$

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  4. #4
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    Thanks for the help. Something is confusing me though:

    Why is the 2 in $\displaystyle \alpha + \beta = -2$ negative?
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  5. #5
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    Quote Originally Posted by Patrick_John View Post
    Let $\displaystyle \alpha,\beta$ be the two solutions of the quadratic equation

    $\displaystyle 3x^2+6x+7=0.$
    Sum of the solutions of quadratic equation $\displaystyle ax^2+bx+c=0$ given by $\displaystyle \alpha+\beta=-\frac{b}{a}$

    Now, applying this to $\displaystyle 3x^2+6x+7=0$, yields $\displaystyle \alpha+\beta=-2$
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  6. #6
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    Hello, Patrick_John!

    Why is the 2 in $\displaystyle \alpha + \beta = -2$ negative?
    It's part of the theory . . .

    Instead of confusing you with symbols, I'll use specific examples
    . . and hope you catch the pattern.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    For example: we are given a cubic equation: .$\displaystyle ax^3 + bx^2 + cx +d \:=\:0$

    Divide by the leading coefficient and insert alternating signs.

    . . $\displaystyle x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} \;=\;0$
    . .+ . . - . . .+ . . -

    Suppose $\displaystyle p,\,q,\,r$ are the roots of the cubic.

    Taking the roots one-at-a-time, their sum is: .$\displaystyle \text{-}\frac{b}{a}$
    . . That is: .$\displaystyle p + q + r \:=\:\text{-}\frac{b}{a}$

    Taking the roots two-at-a-time, their sum is: . $\displaystyle \frac{c}{a}$
    . . That is: .$\displaystyle pq + rq + pr \:=\:\frac{c}{a}$

    Taking the roots three-at-a-time, their sum is: . $\displaystyle \text{-}\frac{d}{a}$
    . . That is: .$\displaystyle pqr\:=\:\text{-}\frac{d}{a}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Suppose we are given a quartic equation: .$\displaystyle ax^4 + bx^3 + cx^2 + dx + e \:=\:0$
    . . with roots $\displaystyle p,\,q,\,r,\,s$.

    Divide by the leading coefficient and insert alternating signs:
    . . $\displaystyle x^4 + \frac{b}{a}x^3 + \frac{c}{a}x^2 + \frac{d}{a}x + \frac{e}{a}\:=\:0$
    . .+ . . - . . .+ . . .-. - .+


    Take the roots . . .

    $\displaystyle \begin{array}{cccc}\text{One-at-a-time:} & p + q + r + s & = & \text{-}\frac{b}{a} \\
    \text{Two-at-a-time:} & ab + ac + ad + bc + bd + cd & = & \frac{c}{a} \\
    \text{Three-at-a-time:} & abc + abd + acd + bcd & = & \text{-}\frac{d}{a} \\
    \text{Four-at-a-time:} & pqrs & = & \frac{e}{a}\end{array}$

    Get it?

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  7. #7
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    Ok I think I understand, but indeed I've never seen these method, and I'd like to learn more about it, could anybody post a bit more theory or suggest me some reading?

    Thanks
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