Question

Solve the following quadratic equations by factorization:

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

Answer 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 6(x² + 12 – 7x + x² + 4 – 5x + x² – 3x + 2) = (x² – 3x + 2)(x² + 12 – 7x) 

⇒ 6(3x² + 18 – 15x ) = x⁴ + 12x² – 7x³ – 3x³ – 36x + 21x² + 2x² + 24 – 14x 

⇒ 18x² – 90x + 108 = x⁴ + 12x² – 7x³ – 3x³ – 36x + 21x² + 2x² + 24 – 14x 

⇒ x⁴ – 10x³ + 17x² + 40x + 84 = 0 

Let P(x) = x⁴ – 10x³ + 17x² + 40x + 84 

At x = -2, (-2)⁴ - 10(-2)³ + 17(-2)² + 40(-2) + 84 

= 16 + 80 + 68 - 80 + 84 

P(x) = 0 therefore, x + 2 is a factor of 

P(x).On dividing P(x) by (x + 2), 

we get x³ - 12x² + 41x - 42 

Let g(x) = x³ - 12x² + 41x - 42, 

P(x) = (x - 2)

g(x) at x =-2

g(x) = 0 therefore, x + 2 is a factor of g(x).

On dividing g(x) by (x + 2), 

we get x² - 14x + 49 

Therefore, 

P(x) = (x - 2)(x - 2)(x² - 14x + 49)

Using, (a - b)² =a² + b² - 2ab, 

we have P(x) = (x + 2)²(x – 7)² 

⇒ (x + 2)²(x – 7)² = 0 

Therefore, possible value of 'x' are -2, -2, and -7, -7