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

Aryan Pandey

New member
Dec 14, 2018
1
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.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
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

$A = (a+b+c)$
$B = -[a(r_1+r_2)+b(r_2+r_3)+c(r_1+r_3)]$
$C = (ar_1r_2+br_2r_3+cr_1r_3)$
 
Last edited:

Greg

Perseverance
Staff member
\(\displaystyle (px-a)(qx-b)+(px-a)(rx-c)+(qx-b)(rx-c)=0\)

\(\displaystyle R=\left(\frac ap,\frac bq\right),\,\left(\frac ap,\frac cr\right),\,\left(\frac bq,\frac cr\right)\)