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TMAT 103. Chapter 7 Quadratic Equations. TMAT 103. § 7.1 Solving Quadratic Equations by Factoring. §7 .1 – Solving Quadratic Equations by Factoring. Quadratic Equation – general form: Key principle – Zero Factor Property: If ab = 0, then either a = 0, b = 0, or both.
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TMAT 103 Chapter 7 Quadratic Equations
TMAT 103 §7.1 Solving Quadratic Equations by Factoring
§7.1 – Solving Quadratic Equations by Factoring • Quadratic Equation – general form: • Key principle – Zero Factor Property: • If ab = 0, then either a = 0, b = 0, or both
§7.1 – Solving Quadratic Equations by Factoring • Solving a Quadratic Equation by Factoring (b 0) • If necessary, write the equation in the form ax2 + bx + c = 0 • Factor the nonzero side of the equation • Using the preceding problem, set each factor that contains a variable equal to zero • Solve each resulting linear equation • Check
§7.1 – Solving Quadratic Equations by Factoring • Examples – Solve the following by factoring x2 – 6x + 8 = 0 2x2 + 9x = 5 x – 2x2 = 0
§7.1 – Solving Quadratic Equations by Factoring • Solving a Quadratic Equation by Factoring (b = 0) • If necessary, write the equation in the form ax2 = c • Divide each side by a • Take the square root of each side • Simplify the result, if possible
§7.1 – Solving Quadratic Equations by Factoring • Examples – Solve the following by factoring 4x2 = 9 16 – x2 = 0
TMAT 103 §7.2 Solving Quadratic Equations by Completing the Square
§7.2 – Solving Quad Equations by Completing the Square • Solving a Quadratic Equation by Completing the Square • The coefficient of the second-degree term must equal (positive) 1. If not, divide each side of the equation by its coefficient • Write an equivalent equation in the form x2 + px = q. • Add the square of ½ of the coefficient of the linear term to each side; that is, (½p)2 • The left side is now a perfect square trinomial. Rewrite the left side as a square • Take the square root of each side • Solve for x and simplify, if possible • Check
§7.2 – Solving Quad Equations by Completing the Square • Examples – Solve the following by completing the square x2 – 6x + 8 = 0 2x2 + 9x = 5 x – 2x2 = 0
TMAT 103 §7.3The Quadratic Formula
§7.3 The Quadratic Formula • The general quadratic equationcan now be solved by completing the square • This will generate a formula that can be used to solve any quadratic equation • x will be written in terms of a, b, and c
§7.3 The Quadratic Formula • Solving a Quadratic Equation using the Quadratic Formula • If necessary, write the equation in the form ax2 + bx + c = 0 • Substitute a, b, and c into the quadratic formula • Solve for x • Check
§7.3 The Quadratic Formula • Examples – Solve the following by using the quadratic formula x2 – 6x + 8 = 0 2x2 + 9x = 5 x – 2x2 = 0
§7.3 The Quadratic Formula • Consider the quadratic formula • The discriminant provides insight into the nature of the solutions • discriminant
§7.3 The Quadratic Formula • Discriminant • If b2 – 4ac > 0, there are 2 real solutions • If b2 – 4ac is also a perfect square they are both rational • If b2 – 4ac is not a perfect square, they are both irrational • If b2 – 4ac = 0, there is only one rational solution • If b2 – 4ac < 0, there are two imaginary solutions • Chapter 14
§7.3 The Quadratic Formula • Examples – How many and what types of solutions do each of the following have? x2 – 2x + 17 = 0 x2 – x – 2 = 0 x2 + 6x + 9 = 0 2x2 + 2x + 14 = 0
TMAT 103 §7.4 Applications
§7.4 Applications • Examples • The work done in Joules in a circuit varies with time in milliseconds according to the formula w = 8t2 – 12t + 20. Find t in ms when w = 16J. • A rectangular sheet of metal 24 inches wide is formed into a rectangular trough with an open top and no ends. If the cross-sectional area is 70 in2, find the depth of the trough.
TMAT 103 §14.1 Complex Numbers in Rectangular Form
§14.1 – Complex Numbers in Rectangular Form • Imaginary Unit • In mathematics, i is used • In technical math, i denotes current, so j is used to denote an imaginary number • Rectangular Form of a Complex Number • a is the real component, and bj is the imaginary component
§14.1 – Complex Numbers in Rectangular Form • Examples – Express in terms of j and simplify
§14.1 – Complex Numbers in Rectangular Form • Powers of j j = j j2 = –1 j3 = –j j4 = 1 j5 = j j6 = –1 j7 = –j j8 = 1 … Process continues • Powers of j evenly divisible by four are equal to 1
§14.1 – Complex Numbers in Rectangular Form • Examples – Express in terms of j and simplify
§14.1 – Complex Numbers in Rectangular Form • Additional Information • Complex numbers are not ordered • “Greater than” and “Less than” do not make sense • Conjugate • The conjugate of (a + bj) is (a – bj)
§14.1 – Complex Numbers in Rectangular Form • Addition and subtraction • Complex numbers can be added and subtracted as if they were 2 ordinary binomials (a + bj) + (c + dj) = (a + c) + (b + d)j (a + bj) – (c + dj) = (a – c) + (b – d)j
§14.1 – Complex Numbers in Rectangular Form • Examples – Perform the indicated operation (1 – 2j) + (3 – 5j) (–3 + 13j) – (4 – 7j) (½ – 11j) – (½ – 4j)
§14.1 – Complex Numbers in Rectangular Form • Multiplication • Complex numbers can be multiplied as if they were 2 ordinary binomials (a + bj)(c + dj) = (ac – bd) + (ad + bc)j
§14.1 – Complex Numbers in Rectangular Form • Examples – Multiply (1 – 2j)(3 – 5j) (–3 + 13j)(4 – 7j) (½ – 11j)(½ – 4j)
§14.1 – Complex Numbers in Rectangular Form • Division • Complex numbers can be divided by multiplying numerator and denominator by the conjugate of the denominator
§14.1 – Complex Numbers in Rectangular Form • Examples – Divide
§14.1 – Complex Numbers in Rectangular Form • Solving quadratic equations with a negative discriminant • 2 complex solutions • Always occur in conjugate pairs • Use quadratic formula, or other techniques
§14.1 – Complex Numbers in Rectangular Form • Examples – Solve using the quadratic formula x2 + x + 1 = 0 x2 + 9 = 0
§14.1 – Complex Numbers in Rectangular Form • Adding complex numbers graphically
§14.1 – Complex Numbers in Rectangular Form • Subtracting complex numbers graphically