Review Homework page # 17 – 19, 30 – 32, 38-40

17) Solution -2 Equation(s):y=-.25x^2-x-1 xy

18 No solution Equation(s):y=x^2-6x+11 xy

19) Solution -3, 4 Equation(s):y=-.5x^2+.5x+6 xy

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 x 2 +3x – 28 = 0 we factor (x+7)(x-4) =0 then solve both factors =0 X + 7 = 0x – 4 = 0 -7 = = +4 X = -7 x = 4 The 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 + _ )

= (x - _ ) ( x - _ ) = (x + _ ) ( x - _ )

So to solve x 2 +3x – 28 = 0 we factor (x+7)(x-4) =0 then solve both factors =0 X + 7 = 0x – 4 = 0 -7 = = +4 X = -7 x = 4 The solution is {-7, 4} You can check your answers by substituting the values for x back into the polynomial *

Solving by factoring 1.5x 2 + 3x = 0 Factor and solve using the Zero Product PrincipalZero Product Principal 2. 16(x – 1) = x(x+8) 2)16x-16=x 2 +8x x 2 - 8x + 16 = 0 (x-4) 2 = 0 x = 4

Factoring = ( _x + _ )( _x + _ ) 1.First blanks have a product of a 2.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)

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 4t 2 + 4t – = -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 – 6t 6.4t 2 +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 4x Each term is a perfect square 2x * 2x = 4x 2 7 * 7 = 49 and there is a subtraction sign bet ween them (2x – 7) (2x +7) or (2x+7)(2x-7)

Perfect square trinomial x 2 + 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 twice 4x x * 2 * (-3) =12 (2x -3) 2

Write an equation given the roots roots are -8, 5 using intercept form y=a(x-x 1 )(x-x 2 ) (x- (-8))(x-5) = 0 x 2 +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 B Distance 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 = 40 2 Simplify, combine like terms, and put it in standard form.

(2(x+4)) 2 + (2x) 2 = 40 2 X = runner B’s speed Runner B’s speed is 12 km/h X+ 4 = runner A’s speed Runner A’s speed is 16 km/h

#2 A picture frame measures 12 cm by 20 cm. 84cm 2 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)=84cm x-40x+4x 2 = 84 4x x +156=0 (divide by 4) x x + 39=0 (x-13)(x-3)=0 x=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 tr aveled 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 th ey are 50 miles apart. Find the spe ed of each train.

Word problem Worksheet practice

Homework Practice worksheet page ( ) #1-43 (odd) #48, 56, 60, 65, 69