+394 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
\(\frac{1}{x_{1}} + \frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}\).
+118687 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.
+118687 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
\(\alpha,\beta,\gamma, \delta\;\;\;then\\ \alpha+\beta+\gamma+ \delta=-\frac{B}{A}\\ \text{The sum of the product of root pairs will be}\quad +\frac{C}{A}\\ \text{The sum of the product of root tripples will be}\quad -\frac{D}{A}\\ and\;\;lastly\\ \alpha\beta\gamma \delta=+\frac{E}{A}\\ \)
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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
