Modulus Inequalities Lecture Notes

Introduction to Modulus Inequalities

• Modulus inequalities involve determining the absolute values of expressions.
• Modulus turns any value into its positive equivalent.
  • E.g., modulus of 5 is 5, and modulus of -5 is also 5.
• Used in various mathematical contexts such as geometry and algebra.

Problem Solving Strategy

• Solve modulus inequalities by squaring both sides of the equation.
• This helps in finding possible values for variables inside the modulus.
• Particularly useful for equations in forms like |x| = a or |x| > a.

Example 1: Basic Equation

• Given: (|x - 2| = 4)
• Objective: Solve for x.
• Method:
  1. Square both sides: ((x - 2)^2 = 16)
  2. Expand: (x^2 - 4x + 4 = 16)
  3. Rearrange to form a quadratic: (x^2 - 4x - 12 = 0)
  4. Factorize: ((x - 6)(x + 2) = 0)
  5. Solutions: (x = 6) or (x = -2)

Example 2: Equation with Modulus on Both Sides

• Given: (|x - 5| = |x + 1|)
• Method:
  1. Square both sides: ((x - 5)^2 = (x + 1)^2)
  2. Expand both: (x^2 - 10x + 25 = x^2 + 2x + 1)
  3. Simplify: (-12x + 24 = 0)
  4. Solve linear equation: (x = 2)

Example 3: Inequality with Greater Than

• Given: (|x - 4| > 3)
• Method:
  1. Square both sides: ((x - 4)^2 > 9)
  2. Expand: (x^2 - 8x + 16 > 9)
  3. Rearrange: (x^2 - 8x + 7 > 0)
  4. Factorize: ((x - 7)(x - 1) > 0)
  5. Determine solution regions:
     • (x > 7) or (x < 1)

Example 4: Complex Inequality with Modulus

• Given: (|x + 1| > 2|x + 3|)
• Method:
  1. Square both sides and expand:
     • Left: ((x + 1)^2)
     • Right: (4(x + 3)^2)
  2. Simplify and rearrange: (-3x^2 - 22x - 35 < 0)
  3. Factorize and find roots:
     • Roots: (x = -5), (x = -\frac{7}{3})
  4. Determine solution interval:
     • (-5 < x < -\frac{7}{3})

Conclusion

• Modulus inequalities can be efficiently solved by squaring both sides.
• Careful expansion and simplification are necessary for correct solutions.
• Graphical sketching can aid in understanding solution intervals, especially for inequalities.