Question

The roots of the equation x²+x-p(p+1)=0 where p is constant are

Answer 1

Answer:

p and -(p+ 1)

Step-by-step explanation:

x^2 + {(p+1)-p}x - p(p+1) = 0

x^2 + (p+1)x -px -p(p+1) = 0

x{x + (p+1)} -p{x + (p+1)} = 0

{x + (p+1)}{x - p} = 0

x = -(p+1) or x = p