Download presentation

1
**LIAL HORNSBY SCHNEIDER**

COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

2
**1.4 Quadratic Equations Solving a Quadratic Equation**

Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant

3
**Quadratic Equation in One Variable**

An equation that can be written in the form where a, b, and c are real numbers with a ≠ 0, is a quadratic equation. The given form is called standard form.

4
**Second-degree Equation**

A quadratic equation is a second-degree equation. This is an equation with a squared variable term and no terms of greater degree.

5
Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both.

6
**Solve Solution: USING THE ZERO-FACTOR PROPERTY Example 1 Standard form**

7
**Solve Solution: USING THE ZERO-FACTOR PROPERTY Example 1**

Solve each equation.

8
Square-Root Property A quadratic equation of the form x2 = k can also be solved by factoring Subtract k. Factor. Zero-factor property. Solve each equation.

9
Square Root Property If x2 = k, then

10
**Square-Root Property That is, the solution of**

Both solutions are real if k > 0, and both are imaginary if k < 0 If k = 0, then this is sometimes called a double solution. If k < 0, we write the solution set as

11
**Solve each quadratic equation. a.**

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. a. Solution: By the square root property, the solution set is

12
**Solve each quadratic equation. b.**

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. b. Solution: Since the solution set of x2 = − 25 is

13
**Solve each quadratic equation. c.**

USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. c. Solution: Use a generalization of the square root property. Generalized square root property. Add 4.

14
**Solving A Quadratic Equation By Completing The Square**

To solve ax2 + bx + c = 0, by completing the square: Step 1 If a ≠ 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Square half the coefficient of x, and add this square to both sides of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution.

15
**Solve x2 – 4x –14 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Step 2 Add 14 to both sides. Step 3 add 4 to both sides. Step 4 Factor; combine terms.

16
**Solve x2 – 4x –14 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 4 Factor; combine terms. Step 5 Square root property. Take both roots. Add 2. Simplify the radical. The solution set is

17
**Solve 9x2 – 12x + 9 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Divide by 9. (Step 1) Add – 1. (Step 2)

18
**Solve 9x2 – 12x + 9 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Factor, combine terms. (Step 4) Square root property

19
**Solve 9x2 – 12x + 9 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Square root property Quotient rule for radicals Add ⅔.

20
**Solve 9x2 – 12x + 9 = 0 by completing the square.**

USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Add ⅔. The solution set is

21
The Quadratic Formula The method of completing the square can be used to solve any quadratic equation. If we start with the general quadratic equation, ax2 + bx + c = 0, a ≠ 0, and complete the square to solve this equation for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. We assume that a > 0.

22
Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are

23
Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

24
**Solve x2 – 4x = – 2 Solution: Here a = 1, b = – 4, c = 2**

USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Write in standard form. Here a = 1, b = – 4, c = 2 Quadratic formula.

25
**The fraction bar extends under – b.**

USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Quadratic formula. The fraction bar extends under – b.

26
**The fraction bar extends under – b.**

USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: The fraction bar extends under – b.

27
**Factor first, then divide.**

USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 – 4x = – 2 Solution: Factor out 2 in the numerator. Factor first, then divide. Lowest terms. The solution set is

28
**Use parentheses and substitute carefully to avoid errors.**

USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x – 4. Solution: Write in standard form. Quadratic formula; a = 2, b = – 1, c = 4 Use parentheses and substitute carefully to avoid errors.

29
**Solve 2x2 = x – 4. Solution: The solution set is Example 6**

USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x – 4. Solution: The solution set is

30
Cubic Equation The equation x3 + 8 = 0 that follows is called a cubic equation because of the degree 3 term. Some higher-degree equations can be solved using factoring and the quadratic formula.

31
**Solve Solution or or SOLVING A CUBIC EQUATION Example 7**

Factor as a sum of cubes. or Zero-factor property or Quadratic formula; a = 1, b = – 2, c = 4

32
**Solve Solution SOLVING A CUBIC EQUATION Example 7 Simplify.**

Simplify the radical. Factor out 2 in the numerator.

33
**SOLVING A CUBIC EQUATION**

Example 7 Solve Solution Lowest terms

34
**Goal: Isolate d, the specified variable.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve for the specified variable. Use when taking square roots. a. Solution Goal: Isolate d, the specified variable. Multiply by 4. Divide by .

35
**See the Note following this example.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. a. Solution Divide by . Square root property See the Note following this example.

36
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. a. Solution Square root property Rationalize the denominator.

37
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. a. Solution Rationalize the denominator. Multiply numerators; multiply denominators.

38
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. a. Solution Multiply numerators; multiply denominators. Simplify.

39
**Solving for a Specified Variable**

Note In Example 8, we took both positive and negative square roots. However, if the variable represents a distance or length in an application, we would consider only the positive square root.

40
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. b. Solution Write in standard form. Now use the quadratic formula to find t.

41
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. b. Solution a = r, b = – s, and c = – k

42
**Solve the specified variable. Use when taking square roots.**

SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use when taking square roots. b. Solution a = r, b = – s, and c = – k Simplify.

43
The Discriminant The Discriminant The quantity under the radical in the quadratic formula, b2 – 4ac, is called the discriminant. Discriminant

44
**The Discriminant Discriminant Number of Solutions Type of Solution**

Positive, perfect square Two Rational Positive, but not a perfect square Irrational Zero One (a double solution) Negative Nonreal complex

45
**Caution The restriction on a, b, and c is important**

Caution The restriction on a, b, and c is important. For example, for the equation the discriminant is b2 – 4ac = = 9, which would indicate two rational solutions if the coefficients were integers. By the quadratic formula, however, the two solutions are irrational numbers,

46
**For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is **

USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a. Solution For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is b2 – 4ac = 22 – 4(5)(– 4) = 84 The discriminant is positive and not a perfect square, so there are two distinct irrational solutions.

47
**First write the equation in standard form as **

USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. b. Solution First write the equation in standard form as x2 – 10x + 25 = 0. Thus, a = 1, b = – 10, and c = 25, and b2 – 4ac =(– 10 )2 – 4(1)(25) = 0 There is one distinct rational solution, a “double solution.”

48
**For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so **

USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. c. Solution For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so b2 – 4ac = (–1)2 – 4(2)(1) = –7. There are two distinct nonreal complex solutions. (They are complex conjugates.)

Similar presentations

Presentation is loading. Please wait....

OK

Break Time Remaining 10:00.

Break Time Remaining 10:00.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google