+0

# Polynomial Equation

+4
134
7

The equation

$$x^4-7x^3+4x^2+7x-2=0$$

has four real roots, $$x_{1}, x_{2}, x_{3}, x_{4}.$$ Find the value of the sum

$$\frac{1}{x_{1}} + \frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}$$.

Jan 2, 2020
edited by hellospeedmind  Jan 3, 2020

#1
+1

1/x_1 + 1/x_2 + 1/x_3 + 1/x_4 = -7.

Jan 2, 2020
#2
+2

This polynomial is degree 4 NOT 5

Let the roots be

$$\alpha,\beta,\gamma,\delta\\ then\\ \alpha+\beta+\gamma+\delta=-\frac{-7}{1}=7\\ \text{The sum of the product of pairs of roots is}\;+\frac{4}{1}=4\\ \text{The sum of the product of triples of roots is}\;-\frac{7}{1}=-7\\ \alpha\beta\gamma\delta=+\frac{-2}{1}=2\\$$

$$\frac{1}{x_{1}} + \frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}\\$$

Now make this one entire fraction and use the facts above to get  a value.

Jan 2, 2020
#3
+1

Hello Melody,

how did you know that a+b+y+o=7??

Guest Jan 2, 2020
#4
+2

Consider the trinomal

$$2x^2-7x-4=0\\ \text{this factorises to }\;\;(2x+1)(x-4)=0\\ \text{so the roots }\alpha\;\;and\;\;\beta \;\; are \;\;\frac{-1}{2}\;\;and \;\;+4\\ \alpha+\beta=\frac{-1}{2}+4=3\frac{1}{2}=\frac{7}{2}\\ \alpha*\beta=\frac{-1}{2}*4=-2=-\frac{4}{2}\\ \\~\\$$

Compare thes numbers to the coefficients in the original equation.

What you can see is not a coincidence.

And it can be extanded to polynomials of ANY degree.

$$Ax^4+Bx^3+Cx^2+Dx^1+E=0$$

If there are 4 real roots and they are

$$\alpha,\beta,\gamma, \delta\;\;\;then\\ \alpha+\beta+\gamma+ \delta=-\frac{B}{A}\\ \text{The sum of the product of root pairs will be}\quad +\frac{C}{A}\\ \text{The sum of the product of root tripples will be}\quad -\frac{D}{A}\\ and\;\;lastly\\ \alpha\beta\gamma \delta=+\frac{E}{A}\\$$

------------------

I am not sure how it works if the roots are not real.  I can't remember thinking about that before.

Here is a video that covers some of this.

Coding:

2x^2-7x-4=0\\
\text{this factorises to }\;\;(2x+1)(x-4)=0\\
\text{so the roots }\alpha\;\;and\;\;\beta \;\; are \;\;\frac{-1}{2}\;\;and \;\;+4\\
\alpha+\beta=\frac{-1}{2}+4=3\frac{1}{2}=\frac{7}{2}\\
\alpha*\beta=\frac{-1}{2}*4=-2=-\frac{4}{2}\\

Ax^4+Bx^3+Cx^2+Dx^1+E=0

Melody  Jan 2, 2020
#5
+1

Thank you, that was really useful!! Guest Jan 2, 2020
#6
+1

You are welcome.  :)

Melody  Jan 2, 2020
#7
+2

Call the roots  A, B, C, D

A + B + C + D =   7

BCD +  ACD  +  ABD  +  ABC   =   -7

ABCD   =  -2

So....we have....

1                1                1                1

___  +    _____    +   _____    +   _____   =

A              B                 C                D

BCD  +  ACD   +   ABD    +   ABC

______________________________   =

ABCD

-7

_____  =        3.5

-2   Jan 2, 2020