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