Holt Algebra 2
5-6 The Quadratic Formula5-6 The Quadratic Formula
Holt Algebra 2
Warm UpLesson PresentationLesson Quiz
Holt Algebra 2
5-6 The Quadratic Formula
Warm Up
Evaluate b2 – 4ac for the given values of the variables.
4. a = 1, b = 3, c = –33. a = 2, b = 7, c = 59 21
2. Solve the equation.3x2 + 96 = 0
1. Express in terms of i.
Holt Algebra 2
5-6 The Quadratic Formula
Solve quadratic equations using the Quadratic Formula.Classify roots using the discriminant.
Objectives
Holt Algebra 2
5-6 The Quadratic Formula
discriminantVocabulary
Holt Algebra 2
5-6 The Quadratic Formula
You have learned several methods for solving quadratic equations: graphing, making tables, factoring, using square roots, and completing the square. Another method is to use the Quadratic Formula, which allows you to solve a quadratic equation in standard form.By completing the square on the standard form of a quadratic equation, you can determine the Quadratic Formula.
Holt Algebra 2
5-6 The Quadratic Formula
Holt Algebra 2
5-6 The Quadratic Formula
To subtract fractions, you need a common denominator.
Remember!
Holt Algebra 2
5-6 The Quadratic Formula
The symmetry of a quadratic function is evident in the last step, . These two zeros are the same distance, , away from the axis of symmetry, ,with one zero on either side of the vertex.
Holt Algebra 2
5-6 The Quadratic Formula
You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.
Holt Algebra 2
5-6 The Quadratic Formula
Holt Algebra 2
5-6 The Quadratic Formula
Find the zeros of f(x)= 2x2 – 16x + 27 using the Quadratic Formula.
Example 1: Quadratic Functions with Real Zeros
Set f(x) = 0.Write the Quadratic Formula.
Substitute 2 for a, –16 for b, and 27 for c.
Simplify.
Write in simplest form.
2x2 – 16x + 27 = 0
Holt Algebra 2
5-6 The Quadratic Formula
Check Solve by completing the square.
Example 1 Continued
Holt Algebra 2
5-6 The Quadratic Formula
Find the zeros of f(x) = x2 + 3x – 7 using the Quadratic Formula.
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 1 for a, 3 for b, and –7 for c.
Simplify.
Write in simplest form.
Check It Out! Example 1a
x2 + 3x – 7 = 0
Holt Algebra 2
5-6 The Quadratic Formula
Check Solve by completing the square.x2 + 3x – 7 = 0
x2 + 3x = 7
Check It Out! Example 1a Continued
Holt Algebra 2
5-6 The Quadratic Formula
Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula.
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 1 for a, –8 for b, and 10 for c.
Simplify.
Write in simplest form.
Check It Out! Example 1b
x2 – 8x + 10 = 0
Holt Algebra 2
5-6 The Quadratic Formula
Check Solve by completing the square.x2 – 8x + 10 = 0
x2 – 8x = –10 x2 – 8x + 16 = –10 + 16
(x + 4)2 = 6
Check It Out! Example 1b Continued
Holt Algebra 2
5-6 The Quadratic Formula
Find the zeros of f(x) = 4x2 + 3x + 2 using the Quadratic Formula.
Example 2: Quadratic Functions with Complex Zeros
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 4 for a, 3 for b, and 2 for c.
Simplify.
Write in terms of i.
f(x)= 4x2 + 3x + 2
Holt Algebra 2
5-6 The Quadratic Formula
Find the zeros of g(x) = 3x2 – x + 8 using the Quadratic Formula.
Set f(x) = 0
Write the Quadratic Formula.
Substitute 3 for a, –1 for b, and 8 for c.
Simplify.
Write in terms of i.
Check It Out! Example 2
Holt Algebra 2
5-6 The Quadratic Formula
The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.
Holt Algebra 2
5-6 The Quadratic Formula
Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!
Holt Algebra 2
5-6 The Quadratic Formula
Find the type and number of solutions for the equation.
Example 3A: Analyzing Quadratic Equations by Using the Discriminant
x2 + 36 = 12xx2 – 12x + 36 = 0b2 – 4ac(–12)2 – 4(1)(36)144 – 144 = 0
b2 – 4ac = 0
The equation has one distinct real solution.
Holt Algebra 2
5-6 The Quadratic Formula
Find the type and number of solutions for the equation.
Example 3B: Analyzing Quadratic Equations by Using the Discriminant
x2 + 40 = 12xx2 – 12x + 40 = 0b2 – 4ac(–12)2 – 4(1)(40)144 – 160 = –16
b2 –4ac < 0
The equation has two distinct nonreal complex solutions.
Holt Algebra 2
5-6 The Quadratic Formula
Find the type and number of solutions for the equation.
Example 3C: Analyzing Quadratic Equations by Using the Discriminant
x2 + 30 = 12xx2 – 12x + 30 = 0b2 – 4ac(–12)2 – 4(1)(30)144 – 120 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.
Holt Algebra 2
5-6 The Quadratic FormulaCheck It Out! Example 3a
Find the type and number of solutions for the equation.
x2 – 4x = –4x2 – 4x + 4 = 0b2 – 4ac(–4)2 – 4(1)(4)16 – 16 = 0
b2 – 4ac = 0
The equation has one distinct real solution.
Holt Algebra 2
5-6 The Quadratic FormulaCheck It Out! Example 3b
Find the type and number of solutions for the equation.
x2 – 4x = –8x2 – 4x + 8 = 0b2 – 4ac(–4)2 – 4(1)(8)16 – 32 = –16
b2 – 4ac < 0
The equation has two distinct nonreal complex solutions.
Holt Algebra 2
5-6 The Quadratic FormulaCheck It Out! Example 3c
Find the type and number of solutions for each equation.
x2 – 4x = 2x2 – 4x – 2 = 0b2 – 4ac(–4)2 – 4(1)(–2)16 + 8 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.
Holt Algebra 2
5-6 The Quadratic FormulaProperties of Solving Quadratic Equations
Holt Algebra 2
5-6 The Quadratic FormulaProperties of Solving Quadratic Equations
Holt Algebra 2
5-6 The Quadratic Formula
No matter which method you use to solve a quadratic equation, you should get the same answer.
Helpful Hint
Holt Algebra 2
5-6 The Quadratic FormulaLesson Quiz: Part I
Find the zeros of each function by using the Quadratic Formula.
1. f(x) = 3x2 – 6x – 5
2. g(x) = 2x2 – 6x + 5
Find the type and member of solutions for each equation.
3. x2 – 14x + 50 4. x2 – 14x + 482 distinct nonreal complex
2 distinct real