roots

[seh2005]

how can i find the sum of the five roots of this equation:
x5 - 4x3 + 5x - 11 = 0

the numbers after the variables (x5) are the exponents, before coefficients...i dont know how to make them superscript

 

[pka]

 

[soroban]

Hello, seh2005!

How can i find the sum of the five roots of this equation:
x⁵ - 4x³ + 5x - 11 = 0

pka is absolutely correct . . . did you understand his hint?
Allow me to baby-talk through the theory . . .

Let's say we're given a cubic equation (degree 3).
If the leading coefficient is not 1, divide through by it
. . and we will have: . x³ + Ax² + Bx + C .= .0

Then I say to myself, "Plus-minus-plus-minus ..." and write them under the coefficients.

. . . x³ + Ax² + Bx + C .= . 0
. + . . - . . .+ . . .-

Suppose the three roots of the cubic are a, b, c.

Take them "one at a time": . a + b + c
. . . This equals -A. . **

Take them "two at a time": . ab + bc + ac
. . . This equals B.

Take them "three at a time": . abc
. . . This equals -C.

This pattern holds for high-degree equations.

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You have a 5th degree equation:
. . . x⁵ + 0x⁴ - 4x³ + 0x² + 5x -11 .= .0
. .+ . . - . . + . . .- . . .+ . . -

If the roots are: a,b,c,d,e
. . . then: .a + b + c + d + e .= .0 . . . There!

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**
In a polynomial equation of degree n, the sum of the n roots
. . is always the negative of the "second coefficient".
(And that is what pka said in his last sentence.)