Question

1/(x-1)(x-2) + 1/(x-2)(x-3) +1/(x-3)(x-4) =1/6

Answer 1

1/(x-1)(x-2) + 1/(x-2)(x-3) + 1/(x-3)(x-4) [simplify--》]
= (4x^2 - 20x + 24)/(x^4 - 10x^3 + 35x^2 - 50x + 24) = 1/6
=》24x^2 - 120x + 144 = x^4 - 10x^3 + 35x^2 - 50x + 24
=》p(x) = x^4 - 10x^3 + 11x^2 + 70x - 120 = 0.By trial, we find x = 2 satisfies the condition.so,(x-2) is a factor of this polynomial.Similarly, (x-3) is its factor.
=》(x-2)(x-3) = x^2 - 5x + 6 = g(x) is also its factor.
Dividing p(x) by g(x), q(x) = x^2 - 5x - 20.
so, x^4 - 10x^3 + 11x^2 + 70x - 120 = (x-2)(x-3)(x^2 - 5x - 20).
Now, solving the polynomial x^2 - 5x - 20
by using quadratic formula,
x = (5+5root5)/2 or (5-5root5)/2.
Hence, x = 2,3,(5+5root5)/2 , (5-5root5)/2.

Answer 2

Answer:

The roots are:

x = 7 or x = -2

Step-by-step explanation:

Given:

$1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)= 1/6$

Take the LCM of the denominators

$(x-1) (x-2) (x-3) (x-4)$

Divide the LCM by the denominator of each term and multiply the result with the  numerator of the respective term.

$[(x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2)] / (x-1)(x-2)(x-3)(x-4) = 1/6$

Use formula:

$(x-a) (x-b) = x^2 - (a+b)x + ab$

After applying the formula:

$(x^2 - 7x + 12 + x^2 - 5x + 4 + x^2 - 3x + 2) / (x-1)(x-2)(x-3)(x-4) = 1/6$

$(3x^2 - 15x + 18) / (x-1)(x-2)(x-3)(x-4) = 1/6\\ \\3(x^2 -5x + 6) / (x -1)(x-2)(x-3)(x-4) = 1/6\\ \\3(x - 2)(x - 3) / (x-1)(x-2)(x-3)(x-4)= 1/6\\ \\3 / (x-1)(x-4)= 1/6\\\\(x-1)(x-4) = 18\\\\x^2 - 5x + 4 = 18\\\\x^2 - 5x - 14 = 0\\\\(x - 7 )(x +2 ) = 0\\\\So,\\x = 7 \\ x = -2$