Problem 6 : Quadratic Equation-2002

( This problem is taken from Indonesian Entrance Test of State University - Year 2002)

If the quadratic equation ( p + 1 ) x^2 - 2 ( p + 3 ) x + 3p = 0 has two equal roots, then the constant p = ...
a. -3 and 3/2 b. -3/2 and 3 c. 1 and 3 d. 2 and -3 e. 3 and -9

Answer :

We know that if the roots are equal then the value of determinant ( D ) = b^2 - 4ac = 0
From the quadratic equation that is given we have a = ( p + 1) ; b = -2 ( p + 3 ) ; c = 3p.
So D = b^2 - 4ac = [ -2(p + 3)]^2 - 4.( p + 1).3p = 0
4 (p + 3)^2 - 12p( p + 1) = 0
4 ( p^2 + 6p + 9) -12p^2-12p = 0
4p^2 + 24p + 36 -12p^2 -12p = 0
-8p^2 + 12 p + 36 = 0
Dividing by -4 we have : 2p ^2 - 3p - 9 = 0
( 2p + 3 )( p - 3) = 0
2p + 3 = 0 or p-3 = 0
2p = -3 p = 3
p = -3/2 Answer : B