CHAPTER 8 Quadratic Equations EQ: What is a quadratic equation?

INTRODUCTION TO QUADRATIC EQUATIONS 8-1 EQ: How are quadratic equations solved by completing the square?

Lesson—Activation Solve by factoring: x(3x + 4) = 0 3x 2 – 6 = 0 14x 2 – 11x = -2

Lesson— Quadratic Equation—a second degree equation Standard Form— ax 2 + bx + c = 0 where a ≠ 0 All quadratic equations must be equal to zero to solve

Lesson—activation 2 Square each of the following: (x + 4) 2 (x – 3) 2 Do you see a pattern?

Try to go in reverse using the rule: x x + 25 Divide the middle term by 2 (x + 5) 2 Try: x 2 – 18x + 81

Lesson What would you need to complete this square: x 2 – 12x + ( x – ) 2

Lesson General rule for completing the square: x 2 + bx + make sure the leading coef =1 x 2 + bx + take half the x term and square it

Lesson How can completing the square assist in solving an equation that can’t be factored? Work with the person next to you to try this problem x 2 + 6x + 7 = 0 x 2 + 6x + = -7+

Lesson—Completing the Square EXAMPLE: 4x 2 – 12x – 7 = 0 standard form 4x 2 – 12x = 7 move the constant x 2 – 3x = 7 divide by the leading 4 coefficient then follow the procedure

HOMEWORK: PAGE(S): 345 – 346 NUMBERS: 3 to 42 by 3’s use factoring up to 16 sq. roots up to 28 completing the square through the end

USING QUADRATIC EQUATIONS 8-2 EQ: How are problems solved by translating into quadratic equations?

Lesson--Activation USE ROPES: which means???? R O P E S Read the problem Organize your thoughts in a chart Plan the equations that will work Evaluate the Solution Summarize your findings

Lesson A rectangular lawn is 60m by 80m. Part of the lawn is torn up to install a pool, leaving a strip of lawn of uniform width around the pool. The area of the pool is 1/6 the area of the old lawn area. How wide is the strip of lawn? x A rect = l w A lawn = 60 x 80 = 4800 A pool = 1/6 A lawn Note: don’t forget to check for extraneous solutions L = ? W =?

Lesson An open box is to be made from 10cm by 20cm rectangular pieces of cardboard by cutting a square from each corner. The area of the bottom of the box is 96 cm 2 What is the length of the sides of the squares that are cut from the corners? x A = lw

HOMEWORK: PAGE(S): 349 NUMBERS: 1,4,5,6, 7,8,9

THE QUADRATIC FORMULA 8-3 EQ: How are quadratic equations solved using the quadratic formula?

Activation: Work with a partner to solve by completing the square: ax 2 + bx + c = 0 ax 2 + bx + = -c x 2 + x + =

Lesson Solve with the quadratic equation: 3x 2 + x - 2 = 0 x 2 – 6x – 20 = 0

Lesson Solve with the quadratic equation: 9x x + 4 = 0 3x 2 + x + 2 = 0

HOMEWORK: Use the quadratic Formula for all problems PAGE(S): 352 NUMBERS: 3 – 30 by 3’s

SOLUTIONS OF QUADRATIC EQUATIONS 8-4 EQ: How are the nature of the solutions determined for a quadratic equation?

ACTIVATION: Look at the homework and compare the problems does there seem to be a pattern that helps you determine the number of solutions?

Lesson The discriminant—(b 2 – 4ac) determines how many solutions and of what type How many solutions do each of the following have? x 2 – 4x + 4 = 0x 2 - 3x – 2 = 0

Lesson Write an equation from the solution If x = 3 or x = -2 what is a possible equation? x – 3 = 0 x + 2 = 0

HOMEWORK: PAGE(S): 357 NUMBERS: 3 – 15 by 3’s and 32, 35, 36

EQUATIONS REDUCIBLE TO QUADRATIC FORM 8-5 EQ: How is substitution used to reduce equations to quadratic form?

ACTIVATION: Given a = 2 b = 3 and c = 5 Find a 2 - 2b + c What procedure did you use?

Lesson U substitution—a method to simplify various equations by placing them in a familiar format example: x 4 – 5x = 0 Let u = __________

Lesson let u = __________ Then

Lesson—you try (x – 1) 2 – 4(x -1) – 5 = 0 Let u = ________

Lesson Let u = _______

HOMEWORK: PAGE(S): 361 NUMBERS: 3 – 24 by 3

FORMULAS AND PROBLEM SOLVING 8-6 EQ: What are quadratic and joint variation?

ACTIVATION: When a ball is on the floor what is its height?

Lesson Given h(t) = -16t 2 + v 0 t + h 0 where h(t) = the height at time t v 0 = the initial velocity h 0 = the initial height When will an object launched at 15 ft/sec from a height of 12 feet will land? How high is it at 1 seconds? The formula is h(t) = -4.9t 2 + v 0 t + h 0 if measurement is given in meters

Lesson Solve for the indicated variable: P = 4s 2 for sA = πr 2 + 2πrh for h

HOMEWORK: PAGE(S): 364 NUMBERS: 3 – 21 by 3’s