If (Px^{2} + (P+1)x + P = 0) has equal roots, find the values of P.
-
A.
(text{-1 and }frac{-1}{3}) -
B.
(text{1 and }frac{-1}{3}) -
C.
(text{-1 and }frac{1}{3}) -
D.
(text{1 and }frac{1}{3})
Correct Answer: Option B
Explanation
For equal roots, (b^{2} – 4ac = 0)
From the equation, (a = P, b = (P+1), c = P)
((P+1)^{2} – 4(P)(P) = P^{2} + 2P + 1 – 4P^{2} = 0)
(-3P^{2} + 2P + 1 = 0 implies 3P^{2} – 2P – 1 = 0)
(3P^{2} – 3P + P – 1 = 0)
(3P(P – 1) + 1(P – 1) = 0)
(P = text{1 or }frac{-1}{3})