MHB Find the roots of f(x)+g(x)+h(x) = 0.

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The discussion centers on finding the roots of the equation f(x) + g(x) + h(x) = 0, where f(x), g(x), and h(x) are quadratic polynomials with positive leading coefficients and real, distinct roots. Each polynomial shares a common root with at least one other, leading to the formulation of f(x), g(x), and h(x) in terms of their roots. The combined polynomial is expressed in standard quadratic form, allowing the application of the quadratic formula to find the roots. The coefficients A, B, and C are derived from the roots and constants associated with each polynomial. The discussion concludes with a focus on the algebraic approach to solving the problem, acknowledging the potential for a more elegant solution.
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Let f(x) , g(x) and h(x) be the quadratic polynomials having positive leading coefficients an real and distinct roots. If each pair of them has a common root , then find the roots of f(x)+g(x)+h(x) = 0.
 
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Aryan Pandey said:
Let f(x) , g(x) and h(x) be the quadratic polynomials having positive leading coefficients an real and distinct roots. If each pair of them has a common root , then find the roots of f(x)+g(x)+h(x) = 0.

Looked at this problem off and on for the last few days trying to come up with some elegant solution. Failed in that endeavor and resorted to grunt work algebra to determine a solution. Please note that this does not mean an "elegant", simple solution doesn't exist.

let $f(x)=a(x-r_1)(x-r_2)$
$g(x) = b(x-r_2)(x-r_3)$,
and $h(x) = c(x-r_1)(x-r_3)$
where $a$, $b$, and $c$ are arbitrary positive constants and $r_1$, $r_2$, and $r_3$ are the real, distinct roots.

$f(x)+g(x)+h(x) = (a+b+c)x^2 - [a(r_1+r_2)+b(r_2+r_3)+c(r_1+r_3)]x+(ar_1r_2+br_2r_3+cr_1r_3)$

quadratic formula ...

$x = \dfrac{-B \pm \sqrt{B^2-4AC}}{2A}$, where

$A = (a+b+c)$
$B = -[a(r_1+r_2)+b(r_2+r_3)+c(r_1+r_3)]$
$C = (ar_1r_2+br_2r_3+cr_1r_3)$
 
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$$R=\left(\frac ap,\frac bq\right),\,\left(\frac ap,\frac cr\right),\,\left(\frac bq,\frac cr\right)$$
 
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