An algebra problem by avn bha

We have the equation

$\dfrac{1}{(x - 1)(x - 2)} + \dfrac{1}{(x - 3)(x - 2)} + \dfrac{1}{(x - 3)(x - 4)} = \dfrac{1}{6}$

Making some adjustment and solving it further:

$\dfrac{(x - 1) - (x - 2)}{(x - 1)(x - 2)} + \dfrac{(x - 2) - (x - 3)}{(x - 3)(x - 2)} + \dfrac{(x - 3) - (x - 4)}{(x - 3)(x - 4)} = \dfrac{1}{6}$

$\dfrac{1}{x - 2} - \dfrac{1}{x - 1} + \dfrac{1}{x - 3} - \dfrac{1}{x - 2} + \dfrac{1}{x - 4} - \dfrac{1}{x - 3} = \dfrac{1}{6}$

$\dfrac{1}{x - 4} - \dfrac{1}{x - 1} = \dfrac{1}{6}$

$18 = (x - 4)(x - 1)$

$x^2 - 5x - 14 =0$

we have the discriminant $\Delta=81\gt 0$ , so both the roots of the quadratic are real and distinct.

By vieta's , sum = $\dfrac{-(-5)}{1} = 5.\square$

Desenvolvendo a equação,

$\frac{(x^2 - 7x + 12) + (x^2 - 5x + 4) + (x^2 - 3x + 2)}{(x - 1)(x - 2)(x - 3)(x - 4)}=\frac{1}{6} \\\\ \frac{3(x^2 - 5x + 6)}{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\\\ \frac{3(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\\\ \frac{3}{(x - 1)(x - 4)} = \frac{1}{6} \\\\ x^2 - 5x + 4 = 18 \\\\ x^2 - 5x - 14 = 0$

Sabemos que a soma das raízes é dada por $- \frac{b}{a}$ ; onde $ax^2 + bx + c = 0$ , $a \neq 0$ . Daí,

$S = - \frac{b}{a} \\\\ S = - \frac{- 5}{1} \\\\ \boxed{S = 5}$

[1/(x- 2)] - [1/(x- 1)] = 1/(x - 1)(x - 2)

[1/(x- 3)] - [1/(x- 2)] = 1/(x - 2)(x - 3)

[1/(x- 4)] - [1/(x- 3)] = 1/(x - 4)(x - 3)

Adding

[1/(x- 4)] - [1/(x-1)] = 1/6

Then

x^2 - 5x - 14 = 0

The two roots are :

7, -2

their sum = 5

or directly

the sum of roots = -(-5)/1 = 5

$\frac{1}{(X - J) * (X - J -1)} =$ $\frac{A}{X - J} + \frac{B}{X - J - 1}$ $\implies$
$A + B = 1$ $(J + 1) * A + J * B = -1$ $\implies$ $A = -1$ $and$ $B = 1$ $\implies$ $\frac{1}{(X - J) * (X - J -1)} =$ $\frac{1}{X - J - 1}$ $-$ $\frac{1} {X - J}$ $\implies$ $\sum\limits_{n = 1}^3 \frac{1}{(X - J) * (X - J -1)}$ $=$ $\frac{1}{X - 4}$ $-$ $\frac{1} {X - 1}$ $=$ $\frac{1}{6}$ $\implies$ $X^2 - 5X - 14 = 0$ $implies$ $X_1,X_2 = \frac{5 +- 9}{2}$ $\therefore$ $X_1 + X_2 = 5$

$In$ $General$ $Solve$ $for$ $X:$ $\sum\limits_{j = 1}^N \frac{1}{(X - J) * (X - J -1)}$ $=$ $\frac{1}{X - N - 1}$ $-$ $\frac{1}{X - 1}$ = $\frac{1}{M}$ $$ $M <> 0.$

$For$ $M > 0$ $we$ $have$ $the$ $real$ $solutions:$

$X_1,X_2 = \frac{(N + 2) \pm \sqrt{N * (N + 4 * M)}}{2}$ $and$ $X_1 + X_2 = N + 2$

take 1/(x-2)as common from the first two terms and then further solve . after that you will observe that (x-2)gets cancelled. then do the same with the next two terms.