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Find the values of p for which the equation x² – (p – 2)x + 2p +1 = 0 has equal roots

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Find the values of p for which the equation x² – (p – 2)x + 2p +1 = 0 has equal roots

A. (1, 2)

B. (21, 0)

 C. (0, 12)

D. (4, 5)

QUICK ANSWER…

C

DETAILS…

For any quadratic equation to have two roots that are exactly equal, b² must be equal to 4ac. Let me explain:

Let’s say a quadratic equation is given as ax² + bx + c = 0, then according to the almighty formula,

Find the values of p for which the equation x² - (p - 2)x + 2p +1 = 0 has equal roots

squaring both sides, b² – 4ac = 0 or b² = 4ac

applying this to solve the question, we have:

x² – (p – 2)x + 2p + 1 = 0

from this,

a = 1

b = -(p – 2)

c = 2p + 1

performing b² = 4ac, we have:

(-(p – 2))² = 4 × 1 × (2p + 1)

RELATED!  Solve the following equation: 2/(2r – 1) – 5/3 = 1/(r + 2)

(p – 2)² = 4(2p + 1)

(p – 2)(p – 2) = 4(2p + 1)

p² – 2p – 2p + 4 = 8p + 4

p² – 4p + 4 = 8p + 4

p² – 4p – 8p + 4 – 4  = 0

p² – 12p = 0

p(p – 12) = 0

therefore, p = 0 or 12

Now for the right answer to the above question:

  1. Option A is incorrect.
  2. Option B is incorrect.
  3. C is correct. 0 actually gives equal roots of 1 and 1, while 12 gives equal roots of 5 and 5.
  4. D is not the correct answer.

KEY-POINTS…

You may please note these/this:

  • For any quadratic equation to have repeated or equal roots, b² must be = 4ac, where a, b, and c are co-efficients according to ax² + bx + c.
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/ culled from 2020 JAMB-UTME mathematics past question 08 /

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