Linear and Quadratic Equations

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If roots are real then the discriminant is >= 0.

Real Roots

    If f(y) = x2 + (2p + 1)x + p2 - 1 and  x is a real number, for what values of ‘p' the function becomes 0?
    1. p > 0
    2. p > -1
    3. p >= -5/4
    4. p <= 3/4

 

  • Correct Answer
    Choice C. p >= -5/4

Detailed Solution

The function f(y) is a quadratic equation.
It is given that x is real.
So the discriminant of f(y) ≥ 0 
i.e. D = b2 - 4ac ≥ 0 or
(2p + 1)2 – 4(p2 - 1) ≥ 0
4p2 + 4p + 1 – 4 (p2 - 1)  ≥ 0
4p + 5 ≥ 0
Or p ≥ - 5/4

Correct Answer: p >= - 5/4



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More Questions from Linear & Quad. Equations

  1. Integer Values; Modulus
  2. Integer Values; Modulus
  3. Integer Values; Modulus
  4. Quadratic Discriminant
  5. Quadratic Discriminant
  6. Quadratics - Counting
  7. Integer Solutions
  8. Linear Equations - Median
  9. Condition for Unique Soln.
Solving equations well is an integral part to cracking any competitive exam. Get cartloads of practice on these two topics.