The quadratic equation $ax^2+bx+c=0$ and $Ax^2+Bx+C$ have roots $(x_1,x_2)$ and $(x_3,x_4)$. Find the condition in which the devide.

Question

The quadratic equations $ax^2+bx+c=0 $ and $Ax^2+Bx+C=0$ have roots $x_1,x_2$ and $x_3,x_4$. Find the condition that the points $(x_1,0)$ and $(x_2,0)$ divide the segment between $(x_3,0)$ and $(x_4,0)$ harmonically. Given two points, $p$ and $q$ are said to divide $a b$ harmonically if one divides $a b$ internally and the other one divides $a b$ externally in the same ratio, in other words, the sum of the ratios is zero. I tried out this question using Section formula and putting the values but I could not get the desired result please help me out.

Answer 1

We can manipulate the relation given by dxiv as follows:

$\frac{x_1-x_3}{x_1-x_4}=\frac{x_3-x_2}{x_2-x_4}$

$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$

Rearranging and summing gives:

$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$

$x_1x_2=\frac{c}{a}$

$x_3x_4=\frac{C}{A}$

$x_3+x_4=-\frac{B}{A}$

$x_1+x_2=-\frac{b}{a}$

Substituting we get:

$2(\frac{c}{a}+\frac{C}{A})-\frac{b}{a}.\frac{B}{A}=0$

Or:

$A\times c+a\times C =\frac{b \times B}{2}$