Lecture Notes: Solving Quadratic Equations Using the Square Root Property

Introduction

• Instructor Name: Fiore
• Objective: To solve quadratic equations using the square root property.

Square Root Property

• Definition:
  • If (u) is an algebraic expression and (d) is a non-zero number, then (u^2 = d) has exactly two solutions.
  • Solutions are (u = \sqrt{d}) or (u = -\sqrt{d}).
  • Alternatively expressed as (u = \pm\sqrt{d}).

Example Solutions Using Square Root Property

Example A

• Equation: (3x^2 - 21 = 0)
• Steps to Solve:
  1. Add 21 to both sides: (3x^2 = 21).
  2. Divide by 3: (x^2 = 7).
  3. Apply square root property: (x = \pm\sqrt{7}).
• Solution Set: ({ \sqrt{7}, -\sqrt{7} }).

Example B

• Equation: (5x^2 + 45 = 0)
• Steps to Solve:
  1. Subtract 45 from both sides: (5x^2 = -45).
  2. Divide by 5: (x^2 = -9).
  3. Apply square root property (complex solution): (x = \pm i\sqrt{9}).
  4. Simplify: (x = \pm 3i).
• Solution Set: ({ 3i, -3i }).

Example C

• Equation: ((x + 5)^2 = 11)
• Steps to Solve:
  1. Apply square root property: (x + 5 = \pm\sqrt{11}).
  2. Subtract 5 from both sides: (x = -5 \pm \sqrt{11}).
• Solution Set: ({ -5 + \sqrt{11}, -5 - \sqrt{11} }).

Practice Problem

• Equation: (4x^2 - 16 = 0)
• Options:
  1. (x = \pm 2) (Correct answer)
  2. (x = 4) (Incorrect)
  3. (x = \pm 4) (Incorrect)

Solution to Practice Problem

• Steps:
  1. Add 16 to both sides: (4x^2 = 16).
  2. Divide by 4: (x^2 = 4).
  3. Apply square root property: (x = \pm 2).

Conclusion

• Key Point: When solving equations using the square root property, remember that (u^2 = d) results in (u = \pm\sqrt{d}).

Note:

• Always simplify the square root if possible.
• Consider complex numbers for negative square roots.