If Px² + (P+1)x + P = 0 has equal roots, find the values of P.

Correct Answer is : 1 and −13{1 and } -1/31 and 3−1

Correct Answer is : 1 and −13{1 and } -1/31 and 3−1

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WAEC Further Mathematics 2014 Paper

Question : 21

Total: 40

Px^{2} + (P+1)x + P = 0

has equal roots, find the values of P.

Solution:

Concept:

A quadratic equation has equal roots when its discriminant is zero:

b^2 - 4ac = 0

Explanation:

Compare

Px^2 + (P+1)x + P = 0

with

ax^2 + bx + c = 0

So,

a = P

b = P+1

, and

c = P

Set the discriminant to zero:

(P+1)^2 - 4(P)(P) = 0

Expand:

P^2 + 2P + 1 - 4P^2 = 0

Simplify:

-3P^2 + 2P + 1 = 0

Multiply by

-1

3P^2 - 2P - 1 = 0

Factorise the quadratic:

3P^2 - 3P + P - 1 = 0

3P(P-1) + 1(P-1) = 0

(3P+1)(P-1) = 0

Therefore,

3P + 1 = 0

P - 1 = 0

P = -\frac{1}{3}

P = 1

Answer:

P = 1

P = -\frac{1}{3}

Option B.

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