Roots of polynomial equation

**arze** · September 14th 2010, 05:32 PM

Find the value of $\sum\alpha^2$ , $\sum\alpha^2\beta^2$ , $\sum\alpha\beta(\alpha+\beta)$ , and $\sum\frac{1}{\alpha}$ for the following case:
$x^4-x^3+2x+3=0$

I started for the first one:
$\sum\alpha^2=\alpha^2+\beta^2+\gamma^2+\delta^2$
$=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\beta\gamma+\gamma\delta+\alpha\delt a+\alpha\gamma+\beta\delta)$
$=(\sum\alpha)^2-2(\sum\alpha\beta)-2(\alpha\gamma+\beta\delta)$
This is as far as I got, the last term is what stumps me.
Thanks!

**Opalg** · September 15th 2010, 12:12 AM

Originally Posted by arze

Find the value of $\sum\alpha^2$ , $\sum\alpha^2\beta^2$ , $\sum\alpha\beta(\alpha+\beta)$ , and $\sum\frac{1}{\alpha}$ for the following case:
$x^4-x^3+2x+3=0$

I started for the first one:
$\sum\alpha^2=\alpha^2+\beta^2+\gamma^2+\delta^2$
$=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\beta\gamma+\gamma\delta+\alpha\delt a+\alpha\gamma+\beta\delta)$
$=(\sum\alpha)^2-2(\sum\alpha\beta)-2(\alpha\gamma+\beta\delta)$
This is as far as I got, the last term is what stumps me.
Thanks!

In problems of this type, the usual convention is that $\sum\alpha\beta$ means the sum of all products of pairs of roots (not just products taken in cyclic order, but all pairs of products). In other words, $\sum\alpha\beta = \alpha\beta+\beta\gamma+\gamma\delta+\alpha\delta+ \alpha\gamma+\beta\delta$ , and $\sum\alpha^2 = \bigl(\sum\alpha\bigr)^2 - 2\sum\alpha\beta$ . With this convention, $\sum\alpha\beta$ is equal to the coefficient of $x^2$ in the quartic equation, so it is equal to 0.

**arze** · September 15th 2010, 03:55 AM

For the second, I worked out the equivalent for cubic equations
$\sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2\alpha\beta\gamma(\sum\alpha)$
Problem is it doesn't work.
$(0)^2-2(3)(-1)=6$ answer is 10.
Thanks!

**Soroban** · September 15th 2010, 06:36 AM

Hello, arze!

The problem is not stated clearly.
And those Greek letters are too hard to type.

I'll do a few of these . . .

$\text}The roots of: }\,x^4 - x^3 + 2x + 3 \:=\:0\,\text{ are }\,a,b,c,d.$

$\text{Find the value of:}$

. . $(A)\;\;\sum\alpha^2$

. . $(B)\;\;\sum\alpha^2\beta^2$

. . $(C)\;\;\sum\alpha\beta(\alpha+\beta)$

. . $(D)\;\;\sum\frac{1}{\alpha}$

From Vieta's Theorem, we have these equations:

. . $\begin{array}{cccccc}a + b + c + d \;=\; 1 & [1] \\<br /> ab + ac + ad + bc + bc + cd \;=\; 0 & [2] \\<br /> abc + abd + acd + bcd \;=\; -2 & [3] \\<br /> abcd \;=\; 3 & [4] \end{array}$

$(A)\;\text{ We want: }\:a^2 + b^2 + c^2 + d^2$

Square [1]: . $(a+b+c+d)^2 \:=\:1^2$

. . $a^2 + 2ab + 2ac + 2ad + b^2 + 2bc + 2bd + c^2 + 2cd + d^2 \;=\;1$

. . $a^2 + b^2 + c^2 + d^2 + 2\underbrace{(ab + ac + ad + bc + bd + cd)}_{\text{This is 0}} \;=\;1$

Therefore: . $a^2+b^2+c^2+d^2 \;=\;1$

$(D)\;\text{ We want: }\:\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d}$

. . $\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \;=\;\frac{\overbrace{bcd + acd + abd + abc}^{\text{This is }-2}}{\underbrace{abcd}_{\text{This is 3}}}$

Therefore: . $\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \;=\;-\frac{2}{3}$

Get the idea?

**Opalg** · September 15th 2010, 07:03 AM

Originally Posted by arze

For the second, I worked out the equivalent for cubic equations
$\sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2\alpha\beta\gamma(\sum\alpha)$
Problem is it doesn't work.
$(0)^2-2(3)(-1)=6$ answer is 10.
Thanks!

You haven't got the formula for $\sum\alpha^2\beta^2$ quite right. It should be $\sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha) +2\alpha\beta\gamma\delta = 0^2 -2(-2)(1) +2(3) = 10.$

**arze** · September 15th 2010, 07:15 PM

Originally Posted by Opalg

You haven't got the formula for $\sum\alpha^2\beta^2$ quite right. It should be $\sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha) +2\alpha\beta\gamma\delta = 0^2 -2(-2)(1) +2(3) = 10.$

So that formula would do for all quartic equations? Because I have another one, finding $\sum\alpha^2\beta^2$ for the equation $x^4+x-1=0$ .
I should get
$(\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha)+2\alpha\btea\ gamma\delta=(o)^2-2(-1)(0)+2(-1)=-2$
answer is 0.
Thanks!

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