Review Homework

page 233-235

# 17 – 19, 30 – 32, 38-40

17) Solution -2

Equation(s):y=-.25x^2-x-1x y-7.0 -6.25-6.0 -4.0-5.0 -2.25-4.0 -1.0-3.0 -0.25-2.0 0-1.0 -0.250 -1.01.0 -2.252.0 -4.03.0 -6.254.0 -9.05.0 -12.256.0 -16.07.0 -20.25

18 No solutionEquation(s):y=x^2-6x+11x y-6.0 83.0-5.0 66.0-4.0 51.0-3.0 38.0-2.0 27.0-1.0 18.00 11.01.0 6.02.0 3.03.0 2.04.0 3.05.0 6.06.0 11.07.0 18.08.0 27.09.0 38.0

19) Solution -3, 4Equation(s):y=-.5x^2+.5x+6x y-6.0 -15.0-5.0 -9.0-4.0 -4.0-3.0 0-2.0 3.0-1.0 5.00 6.01.0 6.02.0 5.03.0 3.04.0 05.0 -4.06.0 -9.07.0 -15.08.0 -22.09.0 -30.0

30) between -6 and -5 between -4 and -3

31) between 0 and 1 between 2 and 3

32) between -3 and 0 between 6 and 9

38-40

4.3 Solving Quadratic Equations by Factoring

Roots• The solutions of a quadratic equation in

standard form are called roots.

• To solve, we use the Zero Product Principle – a product is zero if and only if at least one of the factors is 0

• For any real numbers a and b, if ab=0 then either a=0, b=0 or both=0.

• Roots are also called zeros because they are the x-intercepts so y=0.

• Zero Product Principle – a product is zero if and only if at least one of the factors is 0

So to solve x2 +3x – 28 = 0 we factor(x+7)(x-4) =0 then solve both factors

=0X + 7 = 0 x – 4 = 0 -7 = -7 +4 = +4X = -7 x = 4The solution is {-7, 4}You can check your answers by

substituting the values for x back into the polynomial

Factoring Trinomials

The factors need to add to b and multiply to c(x + _ )(x + _ )

cbxx 2

= (x - _ ) ( x - _ )

= (x + _ ) ( x - _ )

= (x + _ ) ( x - _ )

cbxx 2

cbxx 2

cbxx 2

So to solve x2 +3x – 28 = 0 we factor(x+7)(x-4) =0 then solve both factors

=0X + 7 = 0 x – 4 = 0 -7 = -7 +4 = +4X = -7 x = 4The solution is {-7, 4}You can check your answers by

substituting the values for x back into the polynomial *

Solving by factoring

1. 5x2 + 3x = 0Factor and solve using the

Zero Product Principal

2. 16(x – 1) = x(x+8)

2)16x-16=x2 +8x x2 - 8x + 16 = 0 (x-4)2 = 0 x = 4

Factoring

= ( _x + _ )( _x + _ )1. First blanks have a product of a2. Product of the numbers in the outside

blanks and the product of the numbers in the inside blanks have a sum of b

3. The numbers in the last blanks have a product of c

(another method on next screen)

cbxax 2

Factoring by grouping

1. Multiply ac and find factors of that number that add to b

2. Split the middle term into two terms 3. Factor each half of the polynomial 4t2 + 4t –

15 4. 4 •-15 = -60 (factors are -60,1;60,-1, -30,2;

30,-2; -20,3; 20,-3; -15,4; 15,-4; -12,5; 12,-5; -10,6; 10,-6) Find the two that add to b (+4)

5. Split the middle term +4t into +10t – 6t6. 4t2 +10t / – 6t– 15 – factor each half

7. 2t(2t+5) -3(2t+5) => the terms in the ( ) should match.

8. (2t – 3)(2t + 5) are the factors

Special Cases

• Difference of Two Squares 4x2 - 49

Each term is a perfect square2x * 2x = 4x2

7 * 7 = 49 and there is a subtraction sign between them

(2x – 7) (2x +7) or (2x+7)(2x-7)

• Perfect square trinomialx2 + 6x + 9 twice the square roots of

the product of the a and c term = the b term 2*1*3 = 6

(x+3)2 the square root of each term occurs twice4x2 - 12x + 9 2 * 2 * (-3) =12 (2x -3)2

Write an equation given the roots

roots are -8, 5using intercept form y=a(x-x1)(x-x2)

(x- (-8))(x-5) = 0x2 +3x-40 = y [standard

form]

Solving Problems Using Quadratic Equations

• Read and understand the problem• Develop and carry out a plan• Solve and check

Problem : Runners A and B leave the same point P at right angles. A runs 4 km/h faster than B. After 2 hours they are 40 km apart. Find the speed of each.

Figure out what we need to do. We have two runners running different directions at right angles. Determine what formula to use. We will need to use the Pythagorean Theorem along with d=rt (v=d/t speed = distance / time)Draw a Diagram

A

B

A

BDistance Apart = C

X = runner B’s speed

X+ 4 = runner A’s speed

A’s distance = 2(x+4) 2 hours*speed

B’s distance = 2x 2 hours * speed(2(x+4))2 + (2x)2 = 402

Simplify, combine like terms, and put it in standard form.

(2(x+4))2 + (2x)2 = 402

speed. negative a havenot can wesince extraneous is 16-

12,16

0)12)(16(

01924

01536328

1600464324

40)2()82(

2

2

22

222

xx

xx

xx

xx

xxx

xx

X = runner B’s speed Runner B’s speed is 12 km/hX+ 4 = runner A’s speed Runner A’s speed is 16 km/h

#2 A picture frame measures 12 cm by 20 cm. 84cm2 of the picture shows inside the frame. Find the width of the frame on either side of the picture.

(12-2x)(cm)(20-2x)(cm)=84cm2

240-24x-40x+4x2 = 844x2 - 64x +156=0 (divide by 4)x2 - 16x + 39=0(x-13)(x-3)=0x=13, x = 3, since 12-2(13) or 20-2(13)<0, x=13 is extraneous and the width is 3cm.

 Two cars leave an intersection.  One car travels north; the other travels east. When the car traveling north had gone 24 miles, the distance between the cars was four miles more than three times the distance traveled by the car heading east.  Find the distance between the cars at that time.

Trains A and B leave the same city at the same time headed east and north respectively. Train B travels at 5 mi/h faster than train A. After 2 hours they are 50 miles apart. Find the speed of each train.

Word problemWorksheet practice

Homework

Practice worksheetpage 242 -243 (272-273)

#1-43 (odd) #48, 56, 60, 65, 69