Which equation has exactly one real solution?

To determine which equation has exactly one real solution, we can use the concept of the discriminant from the quadratic formula. The quadratic formula is given by:

x = (-b ± √(b² – 4ac)) / 2a

The discriminant, which is the part b² – 4ac, plays a crucial role in identifying the nature of the roots:

  • If the discriminant is greater than zero, the equation has two distinct real solutions.
  • If the discriminant is equal to zero, the equation has exactly one real solution (a repeated root).
  • If the discriminant is less than zero, the equation has no real solutions (the roots are complex).

Let’s analyze each of the given equations:

  1. a) 4x² – 12x + 9 = 0
    Here, a = 4, b = -12, and c = 9.
    Discriminant = (-12)² – 4(4)(9) = 144 – 144 = 0.
    This equation has exactly one real solution.
  2. b) 4x² + 12x + 9 = 0
    Here, a = 4, b = 12, and c = 9.
    Discriminant = (12)² – 4(4)(9) = 144 – 144 = 0.
    This equation also has exactly one real solution.
  3. c) 4x² – 6x + 9 = 0
    Here, a = 4, b = -6, and c = 9.
    Discriminant = (-6)² – 4(4)(9) = 36 – 144 = -108.
    This equation has no real solutions.
  4. d) 4x² + 6x + 9 = 0
    Here, a = 4, b = 6, and c = 9.
    Discriminant = (6)² – 4(4)(9) = 36 – 144 = -108.
    This equation also has no real solutions.

From the analysis, both equations a) and b) have exactly one real solution. Therefore, the answer to the question is:

Answer: a) 4x² – 12x + 9 = 0 and b) 4x² + 12x + 9 = 0 both have exactly one real solution.

More Related Questions