Question

Question of the day-Set 1
1. How many different values of b exist such that the equation x^2 + bx + b+1 = 0 has integral roots.​

Answer 1

Answer:

-1 & 5 are two values of b

Step-by-step explanation:

How many different values of b exist such that the equation x^2 + bx + b+1 = 0 has integral roots.​

Roots of x² + bx  + b + 1  are

= (-b  ±√(b² - 4b - 4) )/2

-b  ±√(b² - 4b - 4) = 2k  ( where k is integer)

√(b² - 4b - 4) = 2k + b

Squaring both sides

b² - 4b - 4 = 4k² + b² + 4kb

-4b - 4 = 4k² + 4kb

Dividing by 4 both sides

-b - 1 = k² + kb

=> b (k + 1) = -1 - k²

=> b = - (k² + 1)/(k + 1)

we can get values of b by putting values of k as integer

only k = 0 & 1 will give integral value of b  = -1

k = -2 & -3 will give  b = 5

x² + x -1 + 1 = 0

=> x(x + 1) = 0

=> x = 0 , x = -1 ( integral roots)

x² + 5x + 5 + 1 = 0

=> x² + 2x + 3x + 6 = 0

=> (x +2)(x + 3) = 0