Answer :

Given: x2 + 5x – (a2 + a – 6) = 0

Comparing with standard quadratic equation Ax2 + Bx + C = 0


A = 1, B = 5, C = – (a2 + a – 6)


Discriminant D = B2 – 4AC


= (5)2 – 4.1. – (a2 + a – 6)


= 25+ 4a2 + 4a – 24 = 4a2 + 4a + 1


= (2a + 1)2 > 0 Using a2 + 2ab + b2 = (a + b)2


Hence the roots of equation are real.



= (2a + 1)


Roots are given by




x = (a – 2) or x = – (a + 3)


Hence the roots of equation are (a – 2) or x = – (a + 3)


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