Question

find yhe the roots of equation x^2+x-p(p+1),where p is constant

Answer 1

X2 + x - p(p + 1) = 0

Here it is in the form of ax2 + bx + c = 0

:; a = 1 , b = 1 , c = -p2 - p

D = b2 - 4ac

= 1 - 4 * 1 * (-p2 - p)

= 1 + 4p2 + 4p



Now, alpha = - b + root D / 2a

-1 + 2p + 1 /2

= 2p / 2 , = p

and beta = -b - root D / 2a

= -1 -2p - 1 /2

= -2 - 2p /2

-2 (1 + p) / 2

= -1 - p

Therefore the roots are - ( p) and (-1 - p)

Answer 2

Answer:

x² + x - p(p + 1) = 0

=> x² + x - p² - p = 0

=> x² - p² + x - p = 0

=> (x + p)(x - p) + (x - p) = 0

=> (x - p)(x + p + 1) = 0

Either, x - p = 0 or, x + p + 1 = 0

=> x = p and x = - p - 1

Thus the required roots of the given equation are

x = p, - p - 1.