Completing the Square, Imaginary and Complex Numbers

Completing the Square

• The expression x^2 + bx + (\frac{b}{2})^2 can be factored as (x + \frac{b}{2})^2, and x^2 - bx + (\frac{b}{2})^2 is (x - \frac{b}{2})^2. These are called perfect squares.

• Steps to solve a quadratic equation by completing the square:

  1. Subtract c from both sides of the equation: ax^2 + bx = -c
  2. Divide both sides by a to make the coefficient of x^2 equal to 1: x^2 + \frac{b}{a}x = -\frac{c}{a}
  3. Add the square of half the coefficient of x (which is (\frac{b}{2a})^2) to both sides: x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2
  4. Rewrite the left side as a perfect square using the formulas from the first slide. This results in (x + \frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2
  5. Compute the right side by finding a common denominator and adding the numbers.
  6. Solve the obtained equation using the square root property.

Example

• Solve 3x^2 - 4x - 1 = 0 by completing the square.

  1. Add 1 to both sides: 3x^2 - 4x = 1
  2. Divide both sides by 3: x^2 - \frac{4}{3}x = \frac{1}{3}
  3. Add the square of half the coefficient of x to both sides. Half of -\frac{4}{3} is -\frac{2}{3}, and its square is (\frac{-2}{3})^2 = \frac{4}{9}. Thus, x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{1}{3} + \frac{4}{9}
  4. Rewrite the left side as a perfect square: (x - \frac{2}{3})^2 = \frac{1}{3} + \frac{4}{9}
  5. Compute the right side: \frac{1}{3} + \frac{4}{9} = \frac{3}{9} + \frac{4}{9} = \frac{7}{9}. So, (x - \frac{2}{3})^2 = \frac{7}{9}
  6. Solve using the square root property: x - \frac{2}{3} = \pm \sqrt{\frac{7}{9}} = \pm \frac{\sqrt{7}}{3}. Therefore, x = \frac{2}{3} \pm \frac{\sqrt{7}}{3}

  Remarks

• Completing the square is not the most efficient method for solving quadratic equations; the quadratic formula is generally faster.

• The quadratic formula is derived by completing the square on the general quadratic equation ax^2 + bx + c = 0.

• Completing the square is used for techniques in chapter 2.

Imaginary Numbers

• Definition: The imaginary unit i is defined as i = \sqrt{-1}, which means i^2 = -1.
• If a is positive, then \sqrt{-a} = \sqrt{a} \cdot i = i \sqrt{a}.
• Example: \sqrt{-5} = i \sqrt{5}

Rules

• Some rules of radicals that apply to real numbers do not apply to imaginary numbers.
• Example: \sqrt{-9} \cdot \sqrt{-4} \neq \sqrt{(-9) \cdot (-4)}
  • \sqrt{-9} = 3i and \sqrt{-4} = 2i. Thus, \sqrt{-9} \cdot \sqrt{-4} = (3i)(2i) = 6i^2 = 6(-1) = -6
  • If we incorrectly applied the rule, we would get \sqrt{(-9) \cdot (-4)} = \sqrt{36} = 6, which is wrong.

Complex Numbers

• A complex number is a number of the form a + bi, where a and b are real numbers.
• a is called the real part, and b is called the imaginary part.
• Every real number is also a complex number (e.g., 5 = 5 + 0i).

Algebra of Complex Numbers

• Addition/Subtraction: Combine like terms.
  • (5 - 3i) + (-2 + 8i) = (5 - 2) + (-3 + 8)i = 3 + 5i
  • (5 - 3i) - (-2 + 8i) = (5 - (-2)) + (-3 - 8)i = 7 - 11i
• Multiplication: Foil and combine like terms, remembering that i^2 = -1.
  • (2 + 3i)(2 - 3i) = 4 - 6i + 6i - 9i^2 = 4 + 9 = 13

Conjugates

• The conjugate of a complex number a + bi is a - bi.
• If you multiply a number times its conjugate (a + bi)(a - bi), that's like we got in the second example in the previous slides. You're not gonna get any i's don't have to fool there or anything. You can write the answer directly

Division

• To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part from the denominator.

Complex Solutions

• Consider the equation x^2 + 2x + 2 = 0. Here, a = 1, b = 2, and c = 2.
• Using the quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i

Summary

These notes cover completing the square, imaginary numbers, and complex numbers, including definitions, operations, and examples.