# Thread: Values for alpha and beta.

1. ## Values for alpha and beta.

Hello

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

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

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

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

Thanks.

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

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

$(6)^{2}-4(3)(7)=-48$.........yep, negative discriminant, so only complex solutions.

3. Hello, Patrick_John!

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

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

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

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

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

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

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

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

Why is the 2 in $\alpha + \beta = -2$ negative?

5. Originally Posted by Patrick_John
Let $\alpha,\beta$ be the two solutions of the quadratic equation

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

Now, applying this to $3x^2+6x+7=0$, yields $\alpha+\beta=-2$

6. Hello, Patrick_John!

Why is the 2 in $\alpha + \beta = -2$ 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: . $ax^3 + bx^2 + cx +d \:=\:0$

Divide by the leading coefficient and insert alternating signs.

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

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

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

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

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

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

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

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

Take the roots . . .

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

Get it?

7. 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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