Chapter 6 Equations and Inequalities

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Chapter 6 Equations and Inequalities. 6.5 Quadratic Equations. Objectives. Learn the definition of quadratic equation. Multiply two binomials using the FOIL method. Factor trinomials. Solve quadratic equation by factoring. Solve quadratic equation using the quadratic formula - PowerPoint PPT Presentation

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Chapter 6Equations and

Inequalities

6.5Quadratic Equations

Objectives Learn the definition of quadratic equation. Multiply two binomials using the FOIL method. Factor trinomials. Solve quadratic equation by factoring. Solve quadratic equation using the quadratic

formula Use quadratic equations in applications.

Multiplying Two Binomials Using FOIL Method

Binomial: has 2 terms E.g., x + 3, 3x – 5 FOIL Method

Example Multiply: (x + 3)(x + 4)

Solution: F: First terms = x • x = x²

O: Outside terms = x • 4 = 4xI: Inside terms = 3 • x = 3xL: Last terms = 3 • 4 = 12

(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12

Factoring Trinomials Trinomial: -- has 3 terms Note:

Factored Form F O I L Trinomial Form

(x + 3)( x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12

Factoring an algebraic expression:--finding expression that is a product

A Strategy for Factoring x² + bx + c

Factoring a Trinomial in x² + bx + c (cont.)

Factor: x² + 6x + 8.Solution:1. Enter x as the first term of each factor:

x² + 6x + 8 = (x )(x )2. List all pairs of Factors of constant, 8.

3. Try various combinations of these factors. The correct factorization is the one in which the sum of the Outside and Inside products is equal to 6x.

Factors of 8 8, 1 4, 2 -8, -1 -4, -3

Factoring a Trinomial in x² + bx + c (cont.)

Possible factorization o x2 + 6x + 8

Sum of Outside and Inside Products should equal 6x

(x + 8)(x + 1) X + 8x = 9x

(x + 4)(x + 2) 4x + 2x = 6x

(x – 8)(x – 1) -8x – x = -9x

(x – 4)(x – 2) -2x – 4x = -6xThus, x² + 6x + 8 = (x + 4 )(x + 2 ).

Example Factor: 3x² 20x + 28 Solution:1. Factor: 3x² 20x + 28

3x² 20x + 28 = (3x )(x )2. List all pairs of factors of 28. Because the middle

term , 20x, is negative, both factors must be negative. The negative factorizations are: z 1(28), 2(14), 4(7)

3. Try various combinations of these factors:

Example (cont.)Possible Factorizatins of 3x2 – 20x + 28

Sum of Outside and Inside Products:Should Equal -20x

(3x – 1)(x – 28) 84x – x = 85x

(3x – 28)(x – 1) 3x – 28x = 31x

(3x – 2)(x – 14) 42x 2x = 44x

(3x – 14)(x – 2 6x 14x = 20x

(3x – )(x – 7) 21x – 4x = 25x

(3x – 7)(x – 4) 12x 7x = 19x

Thus, 3x² 20x + 28 = (3x 14 )(x 2).Step 4: Verify using the FOIL method:(3x 14)(x 2) = 3x² 6x 14x + 28 = 3x² 20x + 28.

Solving a Quadratic Equation by Factoring

1. Rewrite (in necessary) the equation in the formax2 + bx + c = 0.

2. Factor.3. Let each factor equal to 0.4. Solve the equations in step 3.5. Check the solutions in the original equation.

Example Solve: x² 2x = 35 Solution: x² 2x 35= 0 (x – 7)(x + 5) = 0 Set each factor equal to zero.

x – 7 = 0 or x + 5 = 0 Solve each equation.

x = 7 x = 5 Check:

7² 2· 7 = 35 (5) ² 2(5) = 3549 14 = 3525 +10 = 3535 = 35 35 = 35

Quadratic Formula Given:

ax2 + bx + c = 0 Solve for x.

Quadratic Formula (cont.)

Quadratic Formula Given: ax2 + bx + c = 0, with a ≠ 0 Solution is given by the Quadratic

Formula:

ExampleSolve: 2x² + 9x – 5 = 0Solution: a = 2, b = 9, c = 5

2

2

4

2

9 9 4(2)( 5)

2(2)

9 81 40

4

9 121

4

b b acx

a

x

9 11

49 11 9 11

or 4 4

2 1 20 or 5

4 2 4

x

x x

x x

Example Solve 2x² = 4x + 1 Solution:

2x² 4x – 1 = 0

a = 2, b = 4 and c = 1

4244

48)(164

2(2)1)4(2)(4)(4)(

x2

262

x,2

62x

262

4624

x

Your Turn Solve the following quadratic equations using the

quadratic formula.1. 3x2 + 2x – 3 = 02. 4x2 – 3x = 73. 5x2 – 1x = – 4

Your Turn Use the quadratic formula to solve the following

equations:1. X2 + 3x – 4 = 02. x(x – 4) = 83. 2x2 – 4x – 3 = 04. 9x2 + 12x + 4 = 0

Application: Blood Pressure and Age

A person's normal systolic blood pressure measured in millimeters of mercury depends on his or her age – for men, according to the formula:P = 0.006A² 0.02A + 120Find the age, to the nearest year, of a man whose systolic blood pressure is 125 mm Hg.

Example (cont.)Solution: From the formula, 125 = 0.006A² – 0.02A + 120.Rewriting, we get0.006A² – 0.02A – 5 = 0zwhere a = 0.006, b = 0.02 and c = 52( 0.02) ( 0.02) 4(0.006)( 5)A

2(0.006)

0.02 0.1204

0.0120.02 0.347

A 27(reject)0.012

0.02+0.347A 31

0.012

Therefore, 31 is the approximate age for a man with a normal systolic blood pressure of 125 mm Hg.

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