MM218 - Unit 7 Seminar Topics Imaginary Numbers Complex Numbers Values of in Solving Quadratic Equations by Factoring

Imaginary Numbers The imaginary number i is defined as follows: i = √(-1) and i2 = -1 Definition: For all positive real numbers a, √(-a) = √(-1) √(a) = i √(a) Example 1: √(-16) = i√(16) = 4i Example 2: √(-54) = i√(54) = i√(9*6) = 3i√(6)

Complex Numbers Definition: A number that can be written in the form a+bi where a and b are real numbers, is a complex number. We say that a is the real part and bi is the imaginary part. Example 1: 6 + 2i is complex Example 2: -5 - 7i is complex Example 3: 4i is complex (because we can write it as: 0 + 4i)

Adding/Subtracting Complex Numbers Add (or subtract) the real parts separately from the imaginary parts Example 1: Add 4+2i and 6 + 3i 4 + 2i + 6 + 3i = (4 + 6) + (2i + 3i) = 10 + 5i Example 2: Subtract (4 + 2i) – (6 + 3i) 4 + 2i - 6 - 3i = (4 - 6) + 2i - 3i = -2 - i

Values of in i = i i5 = i i9 = i i2 = -1 i6 = -1 i10 = -1 Example: Evaluate i30 = (i4)7 (i2) = (1)7(-1) = -1

EVALUATE powers of i i0 = 1 Note the pattern, or cycle, i1 = i 1, i, -1 –i, … i2 = -1 i3 = i2 * I = -i RULE: DIVIDE exponent by 4, and the i4 = i2 * i2 = -1 * -1 = 1 remainder will give the power of i, that is, in = ir, where r is the remainder of division by 4. Example: Evaluate i24 = i(24/4) = i(6 R 0) = i0 = 1

Multiplying Complex Numbers Apply the distributive property Example: 2i (5 – 2i) 2i (5 – 2i) = (2i * 5) + (2i * -2i) = 10i + (-4) i2 = 10i + (-4)*(-1) {Replace i2 with -1} = 4 + 10i {WRITE as Complex Number}

Multiplying Complex Numbers Multiply TWO complex numbers, using FOIL Example: (4+2i) * (6 + 3i) F O I L = (4*6) + (4)*(3i) + (2i)*(6) + (2i)*(3i) = 24 + 12i + 12i + 6 i2 = 24 + 24i + 6(-1) {ADD like terms, replace i2 with -1} = 24 + 24i – 6 = 18 + 24i

Dividing Complex Numbers Multiply numerator & denominator by CONJUGATE (May not keep an i in the denominator) Example: (4+i) / (6 + 3i) (4 + i) * (6 - 3i) = (6 + 3i) * (6 - 3i) F O I L 24 - 12i + 6i - 3i2 = 24 – 6i - 3(-1) = 27 - 6i 36 - 18i + 18i - 9i2 36 - 9(-1) 45 = 3(9 - 2i) = 9 - 2i 3*15 15

Quadratic Equation A quadratic equation has this format (standard form): ax2 + bx + c = 0 Where a, b, and c are real numbers, and the value of a cannot be 0 (there MUST BE a square term)

Solving Quadratics by Factoring Given the quadratic in standard form: x2 + 3x + 2 = 0 To solve this equation, we will use the Zero Factor Property (which states that if a product is zero, than one of the two factors must be zero) In symbol form, property states that if a*b = 0 then a = 0 or b = 0

Solving Quadratics by Factoring The steps for using the Zero Factor Property are outlined in your text also 1. Factor, if possible, the quadratic expression that equals zero. 2. Set each factor equal to 0. 3. Solve the resulting equations to find each root. 4. Check each root by substitution.

Solving Quadratics by Factoring x2 + 3x + 2 = 0 GIVEN (x + 1)(x + 2) = 0 FACTOR Now, either x + 1 = 0 or x + 2 = 0 ZERO factor property Hence, x = -1, x = -2 (CHECK both answers back in ORIGINAL equation)

Solving Quadratics by Factoring 4x2 = 9 GIVEN 4x2 – 9 = 0 Write in STANDARD FORM (2x + 3)(2x – 3) = 0 FACTOR 2x + 3 = 0 or 2x -3 =0 Apply ZERO factor property Hence, x = -3/2, x = 3/2 (CHECK answers)

Solving Quadratics by Factoring x2 + 5x + 6 = 0 GIVEN

Solving Quadratics by Factoring x2 - 16 = 0 GIVEN (x + 4)(x - 4) = 0 FACTOR x + 4 = 0 or x - 4 =0 Apply ZERO factor property Hence, x = -4, x = 4

Solving Quadratics by Factoring Example: Solve x2 + 5x - 14=0 Factor the left (x + 7)(x - 2) = 0 Use Zero Property x + 7 = 0 or x - 2 = 0 Solve each x = -7 and x = 2 Check by substituting into original equation: x = -7: x2 + 5x – 14 = 0 x = 2: x2 + 5x – 14 = 0 (-7)2 + 5(-7) – 14 = 0 (2)2 + 5(2) – 14 = 0 49 - 35 -14 =0 4 + 10 – 14 = 0 49 - 49 = 0 14 - 14 = 0 0 = 0 true 0 = 0 true

Solving Quadratics by Factoring Example: Solve 6x2 - 13x – 5 = 0 Factor the left (Outer product, -30 = -10*3, 10*3, -5*6, 5*-6, -15*2, 15*-2) 6x2 - 13x – 5 = 0 6x2 - 15x + 2x – 5 = 0 3x(2x - 5) + 1(2x - 5) = 0 (3x + 1) (2x – 5) = 0 Apply ZERO FACTOR principle, 3x + 1= 0 OR 2x - 5 = 0 AND solve each 3x = -1 2x = 5 x = -1/3 x = 5/2

Practice Exercises