+0  
 
+4
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avatar+355 

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
avatar
+1

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

 Jan 2, 2020
 #2
avatar+108732 
+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
avatar
+1

Hello Melody,

how did you know that a+b+y+o=7?? 

Guest Jan 2, 2020
 #4
avatar+108732 
+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.

 

So in your polynomial

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

 

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.

https://www.khanacademy.org/math/math-for-fun-and-glory/aime/2003-aime/v/sum-of-polynomial-roots-proof

 

 

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
avatar
+1

Thank you, that was really useful!! smiley

Guest Jan 2, 2020
 #6
avatar+108732 
+1

You are welcome.  :)

Melody  Jan 2, 2020
 #7
avatar+109563 
+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

 

 

 

cool cool cool

 Jan 2, 2020

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