Complex Numbers and Quadratic Equations

Real Number System and Complex Numbers

  • Real Numbers (RR):

    • The term real numbers represents all numbers on the real number line.
    • The range of real numbers extends from negative infinity (-\infty) to positive infinity (++\infty).
    • This set includes all decimals, fractions, whole numbers, zero, positive numbers, and negative numbers.
  • Complex Numbers:

    • The complex number system is the set of all numbers that take the form a+bia + bi.
    • In this form, aa and bb must be real numbers.
    • The component ii is defined as the imaginary unit or the imaginary number.
    • Fundamental Equations and Definitions:
      • i=1i = \sqrt{-1}
      • i2=1i^2 = -1
    • Standard Form:
      • Any complex number should be expressed in the form a+bia + bi.
      • Example examples including 5+2i5 + 2i or 73i7 - 3i.
      • If the real part (aa) or the imaginary part (bibi) involves a zero, it is typically omitted in writing (e.g., 0+2i0 + 2i is written simply as 2i2i).

Operations with Complex Numbers

  • Addition and Subtraction:

    • Complex numbers are added or subtracted similarly to binomials, treating ii like a variable such as xx.
    • The procedure involves grouping and combining common terms (real parts with real parts, imaginary parts with imaginary parts).
    • Addition Example: (52i)+(3+3i)(5 - 2i) + (3 + 3i)
      • Combine real parts: 5+3=85 + 3 = 8
      • Combine imaginary parts: 2i+3i=1i-2i + 3i = 1i (or just ii)
      • Final result in standard form: 8+i8 + i
    • Subtraction Example: (12+6i)(2i)(12 + 6i) - (2 - i)
      • Subtract the real parts: 122=1012 - 2 = 10
      • Subtract the imaginary parts: 6i(i)6i - (-i). Subtracting a negative is equivalent to addition (6i+i=7i6i + i = 7i).
      • Final result in standard form: 10+7i10 + 7i
  • Multiplication:

    • Distributive properties are applied to multiply complex numbers.
    • Single Term Distribution Example: 7i(29i)7i(2 - 9i)
      • Multiply 7i7i by 22 to get 14i14i.
      • Multiply 7i7i by 9i-9i to get 63i2-63i^2.
      • Substitute 1-1 for every instance of i2i^2: 14i63(1)=14i+6314i - 63(-1) = 14i + 63.
      • Rewrite in standard form: 63+14i63 + 14i.
    • Binomial Multiplication (FOIL Method):
      • FOIL stands for First, Outside, Inside, Last.
      • Example: (5+4i)(67i)(5 + 4i)(6 - 7i)
        1. First: 5×6=305 \times 6 = 30
        2. Outside: 5×(7i)=35i5 \times (-7i) = -35i
        3. Inside: 4i×6=24i4i \times 6 = 24i
        4. Last: 4i×(7i)=28i24i \times (-7i) = -28i^2
      • Combine and simplify: 3035i+24i28(1)=3011i+28=5811i30 - 35i + 24i - 28(-1) = 30 - 11i + 28 = 58 - 11i.

Conjugates and Division of Complex Numbers

  • Conjugate of a Complex Number:

    • The conjugate is formed by changing the operation between the real and imaginary components to the opposite sign.
    • The conjugate of a+bia + bi is abia - bi.
    • The conjugate of abia - bi is a+bia + bi.
  • Dividing Complex Numbers:

    • To divide complex numbers, multiply both the numerator and the denominator by the conjugate of the denominator.
    • Example Process: 5+4i4i\frac{5 + 4i}{4 - i}
      • Multiply by the conjugate of the denominator: 5+4i4i×4+i4+i\frac{5 + 4i}{4 - i} \times \frac{4 + i}{4 + i}
      • Numerator Distribution: 5(4)+5(i)+4i(4)+4i(i)=20+5i+16i+4i2=20+21i4=16+21i5(4) + 5(i) + 4i(4) + 4i(i) = 20 + 5i + 16i + 4i^2 = 20 + 21i - 4 = 16 + 21i
      • Denominator Distribution: 4(4)+4(i)i(4)i(i)=16+4i4ii2=16(1)=174(4) + 4(i) - i(4) - i(i) = 16 + 4i - 4i - i^2 = 16 - (-1) = 17
      • Result: 16+21i17\frac{16 + 21i}{17}
      • Standard Form Conversion: Splitting the fraction results in 1617+2117i\frac{16}{17} + \frac{21}{17}i.
  • Practice Problem: 3i7+i\frac{3i}{7 + i}

    • Multiply by conjugate: 3i(7i)(7+i)(7i)\frac{3i(7 - i)}{(7 + i)(7 - i)}
    • Numerator: 21i3i2=21i3(1)=3+21i21i - 3i^2 = 21i - 3(-1) = 3 + 21i
    • Denominator: 497i+7ii2=49(1)=5049 - 7i + 7i - i^2 = 49 - (-1) = 50
    • Result: 3+21i50\frac{3 + 21i}{50}

Square Roots of Negative Numbers

  • Core Principle:

    • If there is a negative number under a square root, remove the negative sign and place the imaginary unit (ii) outside of the square root.
    • Example: 44i2i\sqrt{-4} \rightarrow \sqrt{4}i \rightarrow 2i
  • Simplifying Square Roots:

    • Break down the number into its prime factors.
    • Look for pairs of factors. For every pair, one representative of that factor is pulled outside the square root.
    • Example: 27+48\sqrt{-27} + \sqrt{-48}
      • Convert to imaginary forms: 27i+48i\sqrt{27}i + \sqrt{48}i
      • Factoring 2727: 3×3×3=32×33 \times 3 \times 3 = 3^2 \times 3. Pull out the 33 to get 33i3\sqrt{3}i.
      • Factoring 4848: 4×12=4×4×3=42×34 \times 12 = 4 \times 4 \times 3 = 4^2 \times 3. Pull out the 44 to get 43i4\sqrt{3}i.
      • Combine like terms: 33i+43i=73i3\sqrt{3}i + 4\sqrt{3}i = 7\sqrt{3}i.
  • Additional Example: 84+5498\sqrt{-4} + 5\sqrt{-49}

    • Simplify: 8(4i)+5(49i)8(\sqrt{4}i) + 5(\sqrt{49}i)
    • Calculate: 8(2i)+5(7i)=16i+35i=51i8(2i) + 5(7i) = 16i + 35i = 51i

Quadratic Functions and Equations

  • Definition:

    • A quadratic function is any function where the highest power of the variable (xx) is 22.
    • Standard form: ax2+bx+c=0ax^2 + bx + c = 0.
    • Also referred to as a second-degree polynomial.
  • Zero Product Property:

    • If a×b=0a \times b = 0, then either a=0a = 0 or b=0b = 0.
    • This property is essential for solving quadratic equations after factoring.
  • General Steps for Solving Quadratic Equations:

    1. Rewrite the equation in standard form (ax2+bx+c=0ax^2 + bx + c = 0).
    2. Factor the quadratic completely.
    3. Set each individual factor equal to zero using the zero product property.
    4. Solve the resulting linear equations for the variable.
    5. Check solutions against the original equation.

Factoring Strategies and Examples

  • Standard Factoring Example: 2x2+x=12x^2 + x = 1

    1. Standard form: 2x2+x1=02x^2 + x - 1 = 0
    2. AC Method: a×c=2×(1)=2a \times c = 2 \times (-1) = -2. Factors of 2-2 that add to get +1+1 are +2+2 and 1-1.
    3. Group: (2x2+2x)(1x+1)=0(2x^2 + 2x) - (1x + 1) = 0
    4. Factor by grouping: 2x(x+1)1(x+1)=0(2x1)(x+1)=02x(x + 1) - 1(x + 1) = 0 \rightarrow (2x - 1)(x + 1) = 0
    5. Solve: 2x1=0x=1/22x - 1 = 0 \rightarrow x = 1/2; x+1=0x=1x + 1 = 0 \rightarrow x = -1
  • Factoring by Grouping (Missing C term): 4x22x=04x^2 - 2x = 0

    1. Factor out the greatest common factor (GCF): 2x(2x1)=02x(2x - 1) = 0
    2. Set factors to zero: 2x=02x = 0 and 2x1=02x - 1 = 0
    3. Solutions: x=0x = 0 and x=1/2x = 1/2
  • Square Root Property:

    • Applicable when there is no linear component (bxbx term, where b=0b = 0).
    • Example: 5x2=455x^2 = -45
      1. Isolate x2x^2: x2=9x^2 = -9
      2. Take the square root of both sides. Crucial: When taking a square root across an equals sign, always add a ±\pm (plus or minus) sign.
      3. Solve: x=±9=±3ix = \pm \sqrt{-9} = \pm 3i

Completing the Square

  • This method is used when quadratic equations cannot be factored traditionally. It is a necessary skill for moving into advanced mathematics like precalculus.

  • Step-by-Step Procedure:

    1. Ensure the leading coefficient (aa) is exactly 11.
    2. Isolate the constant term (cc) by moving it to the opposite side of the equation.
    3. Identify the middle term coefficient (bb) and divide it by 22 (b2\frac{b}{2}).
    4. Square the result from step 3: (b2)2(\frac{b}{2})^2.
    5. Add this squared value to both sides of the equation.
    6. Factor the left side (it will always form a perfect square: (x+b2)2(x + \frac{b}{2})^2) and solve using the square root property.
  • Example: x2+4x1=0x^2 + 4x - 1 = 0

    1. Move the constant: x2+4x=1x^2 + 4x = 1
    2. Determine the value to add: b=4b = 4. Half is 22. Squared is 44.
    3. Add value to both sides: x2+4x+4=1+4x2+4x+4=5x^2 + 4x + 4 = 1 + 4 \rightarrow x^2 + 4x + 4 = 5
    4. Factor as a perfect square: (x+2)2=5(x + 2)^2 = 5
    5. Apply square root property: x+2=±5x + 2 = \pm \sqrt{5}
    6. Final Solution: x=2±5x = -2 \pm \sqrt{5}

The Quadratic Formula

  • The quadratic formula is a universal method that always works for solving quadratic equations in the form ax2+bx+c=0ax^2 + bx + c = 0.
  • Formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • Example Application: 9x2+6x=49x^2 + 6x = 4
    1. Standard form: 9x2+6x4=09x^2 + 6x - 4 = 0 where a=9,b=6,c=4a = 9, b = 6, c = -4
    2. Plug into formula: x=6±624(9)(4)2(9)x = \frac{-6 \pm \sqrt{6^2 - 4(9)(-4)}}{2(9)}
    3. Simplify under the root (the discriminant\text{the discriminant}): 36(144)=18036 - (-144) = 180
    4. Simplified equation: x = \frac{-6 \pm \sqrt{180}}{18}
    5. Factor 180\sqrt{180}: 180=2×90=2×2×45=2×2×9×5=2×2×3×3×5180 = 2 \times 90 = 2 \times 2 \times 45 = 2 \times 2 \times 9 \times 5 = 2 \times 2 \times 3 \times 3 \times 5. Pairs of 22 and 33 result in pulling out a 66 (2×32 \times 3). 180=65\sqrt{180} = 6\sqrt{5}.
    6. Simplify the overall fraction: x=6±6518x=6(1±5)6×3x = \frac{-6 \pm 6\sqrt{5}}{18} \rightarrow x = \frac{6(-1 \pm \sqrt{5})}{6 \times 3}
    7. Final result: x=1±53x = \frac{-1 \pm \sqrt{5}}{3}

The Discriminant

  • The discriminant is the expression found inside the square root of the quadratic formula: b24acb^2 - 4ac.

  • It identifies the nature and number of solutions for a quadratic equation:

    • If b24ac>0b^2 - 4ac > 0 (Positive): There are two real solutions.
    • If b24ac<0b^2 - 4ac < 0 (Negative): There are two complex (imaginary) solutions.
    • If b24ac=0b^2 - 4ac = 0: There is exactly one real solution.
  • Example 1: 8x2+2x1=08x^2 + 2x - 1 = 0

    • b24ac=(2)24(8)(1)=4+32=36b^2 - 4ac = (2)^2 - 4(8)(-1) = 4 + 32 = 36.
    • Value is positive, so there are two real solutions.
  • Example 2: x22x+17=0x^2 - 2x + 17 = 0

    • b24ac=(2)24(1)(17)=468=64b^2 - 4ac = (-2)^2 - 4(1)(17) = 4 - 68 = -64.
    • Value is negative, so there are two complex solutions.

Questions & Discussion

  • Question on Subtraction Sign Change: When subtracting complex numbers such as 6i(i)6i - (-i), why does it become +7i+7i?

    • Response: Subtracting a negative value is mathematically equivalent to addition. Therefore, 6i(i)6i - (-i) becomes 6i+i6i + i, resulting in 7i7i.
  • Question on Solving for xx: How was the solution x=1/2x = 1/2 obtained from 2x1=02x - 1 = 0?

    • Response: First, add 11 to both sides of the equation to get 2x=12x = 1. Then, divide both sides by 22 to isolate xx, yielding x=1/2x = 1/2.
  • Question on Calculator Use: Can students use phones as calculators during tests?

    • Response: Scientific calculators are permitted and encouraged. Phones may be used for calculations during practice or homework, but they are strictly prohibited during exams.
  • Question on Simplification Pickiness: Will the computer homework system accept answers not in standard form?

    • Response: While an instructor might grant partial credit on a written exam for a non-standard form like 14i+6314i + 63, digital homework platforms are very strict and generally require the standard form (63+14i63 + 14i) and the splitting of fractional answers into separate terms (16/17+21/17i16/17 + 21/17i).