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If the equation px² + (p+1)x + p = 0,has equal roots, find the value of p

User Alphy
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Final answer:

The value of p for which the quadratic equation px² + (p+1)x + p = 0 has equal roots is determined by setting the discriminant to zero and solving for p. Upon calculation, the value of p is found to be -1/3.

Step-by-step explanation:

If the quadratic equation px² + (p+1)x + p = 0 has equal roots, the discriminant of the equation must be zero. The discriminant (Δ) is given by Δ = b² - 4ac. For the given equation, a = p, b = p+1, and c = p.

Setting the discriminant to zero, we get:

Setting the discriminant to zero, we get:

(p+1)² - 4(p)(p) = 0

This simplifies to:

p² + 2p + 1 - 4p² = 0

-3p² + 2p + 1 = 0

Factoring or using the quadratic formula:

p = (-2 ± √((2)² - 4(-3)(1))) / (2(-3))

p = (-2 ± √(4 + 12)) / (-6)

p = (-2 ± √(16)) / (-6)

p = (-2 ± 4) / (-6)

Since we need a real value for p that makes sense in the context of the equation, we use:

p = (-2 + 4) / (-6)

p = 2 / (-6)

p = -1/3

Thus, the value of p is -1/3 when the equation has equal roots.

User MickG
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