Sum of roots

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SuperDude

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What is the sum of the roots of the equation 8x<sup>3</sup> - 2x<sup>2</sup> + 8x - 7 = 0

I believe the answer that they gave was 1/4, but how do we get there? Is there some formula that I should know that would allow me to easily find the sum of the roots of a polynomial? I know I can easily find the sum of the roots of a second degree polynomial, but a third?

TIA.
 
G'day,

This is something I learned from Matt...

The sum of the roots of a cubic
ax^3 + bx^2 + cx + d = 0

is given by -b/a

The same goes for any polynomial where you look at the highest and second-highest powers of x in the polynomial.

And the product of the roots of a cubic is given by -d/a.
The sign is negative if the polynomial is of an odd order, positive if even (and you're looking at the "last" and "first" terms.)
 
Hello, SuperDude!

Unco is absolutely correct . . . and there's more . . .

What is the sum of the roots of the equation \(\displaystyle 8x^3\,-\,2x^2\,+\,8x\,-\,7\:=\:0\)
Click to expand...
Given any cubic equation: .\(\displaystyle ax^3\,+\,bx^2\,+\,cx\,+\,d\:=\:0\)

Divide by the leading coefficient: .\(\displaystyle x^3\,+\,\frac{b}{a}x^2\,+\,\frac{d}{a}x\,+\,\frac{d}{a}\:=\:0\)

We have four coefficients: .\(\displaystyle \L1,\;\frac{b}{a},\;\frac{c}{a},\;\frac{d}{a}\)
. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \uparrow\;\,\uparrow\;\uparrow\;\:\uparrow\)
Assign alternating signs: . .+ . .- . .+ . .-

And we have: .\(\displaystyle \L1,\;-\frac{b}{a},\;\frac{c}{a},\;-\frac{d}{a}\)

If \(\displaystyle p,\,q,\,r\) are the roots of the cubic, then:
. . . . . \(\displaystyle \L p\,+\,q\,+\,t\;\;\;=\,-\frac{b}{a}\) . . . . . sum of the roots, taken one at a time
. . . \(\displaystyle \L pq\,+\,qr\,+\,pr\:=\;\;\frac{c}{a}\) . . . . .sum of the roots, taken two at a time
. . . . . . . . . \(\displaystyle \L pqr\;\;\;\;\;=\,-\frac{d}{a}\) . . . . . sum of the roots, taken three at a time

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

This prodecure can expanded to higher-degree polynomial equations.

For example, a quartic: .\(\displaystyle ax^4\,+\,bx^3\,+\,cx^2\,+\,dx\,+\,e\:=\:0\)

If \(\displaystyle p,\,q,\,r,\,s\) are the roots of the quartic, then:
. . . . . . . . . . . . \(\displaystyle \L p\,+\,q\,+\,r\,+\,s\;\;\;\;\;\;\;\;\;=\,-\frac{b}{a}\) . . . . . one at a time
. . . \(\displaystyle \L pq\,+\,pr\,+\,ps\,+\,qr\,+\,qs\,+\,rs\:=\;\;\,\frac{c}{a}\) . . . . . two at a time
. . . . . . .\(\displaystyle \L pqr\,+\,pqs\,+\,prs\,+\,qrs\;\;\;\;=\,-\frac{d}{a}\) . . . . . three at a time
. . . . . . . . . . . . . . . . . .\(\displaystyle \L pqrs\;\;\;\;\;\;\;\;\;\;\;\;\;=\;\;\,\frac{e}{a}\) . . . . . four at a time
 
If you're curious, here's one that is interesting and, who knows, may prove useful one day.

Adding up the reciprocals of all the roots of a polynomial results in the negative

of the ratio of the linear coeffcient to the constant coeffcient.

If \(\displaystyle (x-r_1)(x-r_2).....(x-r_n)=

x^{n}+a_{n-1}x^{n-1}+.......+a_1x+a_0\), then

\(\displaystyle \frac{1}{r_1}+\frac{1}{r_2}+........+\frac{1}{r_n}=\frac{-a_1}{a_0}\)

Here's an example:

\(\displaystyle 3x^{4}+14x^{3}+14x^{2}-8x-8=0\)

This can be expressed as:

\(\displaystyle (x+2)(x+\frac{2}{3})(x-(-1-\sqrt{3}))(x-(-1+\sqrt{3}))\)

The roots are:

\(\displaystyle [-2, \frac{-2}{3}, -1+\sqrt{3}, -1-\sqrt{3}]\)

Adding up the reciprocals of these roots arrives you at -1

Which is the negative of \(\displaystyle \frac{-8}{-8}=1\)

Euler used this fact in proving his famous Basel problem. Pretty cool, huh?.
 
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