Solving Quadratic Solving Quadratic Word Problems by Word Problems by

FactoringFactoring

Solving Quadratic Solving Quadratic Word Problems by Word Problems by

FactoringFactoringMarch 11, 2014March 11, 2014

SWBAT Solve Quadratic Word Problems by Factoring

SWBAT Solve Quadratic Word Problems by Factoring

WARM – UP9 cm 15 cm1. Solve for the missing side.

2.Find two consecutive odd integers such that twice the larger is seven less than three times the smaller.

a2 + b2 =c292 + x2 = 15281 + x2 = 225 x2 = 144 x = 12Let x = 1st integerx + 2 = 2nd odd integer 2(x + 2) = 3x – 72x + 4 = 3x – 7 11 = x 13 = x + 2

SWBAT Solve Quadratic Word Problems by Factoring

STEPS IN SOLVING WORD PROBLEMS 1.Define the variables that you want to find with let statements.2.Create equation(s) that express the information given in the problem’s scenario.3. Solve using algebraic methods.4. Consider if your answer(s) is/are reasonable.5. Label your solution(s) appropriately.6.Check your answer(s) with the conditions given in the problem.  

  

SWBAT Solve Quadratic Word Problems by Factoring

Example 1 The product of two consecutive even integers is 48. Find the integers.  Let x = the smaller integer x + 2 = the larger integer

x(x + 2) = 48x2 + 2x = 48x2 + 2x – 48 = 0(x + 8)(x – 6) = 0x + 8 = 0 x – 6 = 0 x = -8 x = 6 x + 2 = -6 x + 2 = 8 Solutions: {-8, -6} & {6, 8}

SWBAT Solve Quadratic Word Problems by Factoring

Example 2 Find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7 times the second.  

Let x = 1st positive consecutive integer x + 2 = 2nd positive consecutive integer x + 4 = 3rd positive consecutive integerx(x + 4) = 7(x + 2) – 4x 2 + 4x = 7x + 14 – 4x 2 + 4x = 7x + 10x 2 - 3x – 10 = 0(x – 5) (x + 2) = 0x – 5 = 0 x + 2 = 0x = 5 x = -2 REJECTx + 2 = 7 x + 4 = 9{5, 7, 9}

SWBAT Solve Quadratic Word Problems by Factoring

Practice:Find three consecutive positive integers such that the product of the first two is 22 less than 11 times the third. Let x = 1st integer x + 1 = 2nd integer x + 2 = 3rd integerx(x + 1) = 11(x + 2) – 22x2 + x = 11x + 22 – 22x2 – 10x = 0x(x – 10) = 0x = 0 x – 10 = 0 x = 10 x + 1 = 11 x + 2 = 12

{10, 11, 12}

SWBAT Solve Quadratic Word Problems by Factoring

2. The product of two consecutive odd integers is equal to 30 more than the first. Find the integers. 

Let x = 1st integer x + 2 = 2nd odd integerx(x + 2) = x + 30x2 + 2x = x + 30x2 + x – 30 = 0(x + 6)(x – 5) = 0x + 6 = 0 x – 5 = 0 x = -6 x = 5 x + 2 = 7 {5, 7}

SWBAT Solve Quadratic Word Problems by Factoring

Example 3An object is moving in a straight line. It initially travels at a speed of 9 meters per second, and it speeds up at a constant rate of 2 meters per second each second. Under such conditions, the distance d, in meters, that the object travels is given by the equation d = t2 + 9t, where t is in seconds. According to this equation, how long will it take the object to travel 22 meters? 

d = t2 + 9t22 = t2 + 9tt2 + 9t – 22 = 0(t + 11)(t – 2) = 0t + 11 = 0 t – 2 = 0t = -11 REJECT t = 2 seconds

SWBAT Solve Quadratic Word Problems by Factoring

Practice: An object is moving in a straight line. It initially travels at a speed of 6 meters per second, and it speeds up at a constant acceleration of 4 meters per second each second. The distance d in meters, that this object travels is given by the equation d = 2t2 + 6t, where t is in seconds. According to this equation, how long will it take the object to travel 108 meters?  

2t2 + 6t = 1082t2 + 6t – 108 = 02(t2 + 3t – 54) = 0 2(t + 9)(t – 6) = 0 t + 9 = 0 t – 6 = 0 t = -9 t = 6 sec.

SWBAT Solve Quadratic Word Problems by Factoring

 Example 4 One leg of a right triangle is one inch shorter than the other leg. If the hypotenuse is 5 inches, find the length of the shorter leg.x

x - 15 in.

a2 + b2 =c2x2+ (x – 1)2 = 52x2 + x2 - 2x +1 = 252x2 - 2x – 24 = 02(x2 - x – 12) = 02(x - 4)(x + 3) = 0 x - 4 = 0 x + 3 = 0 x = 4 x = - 3 x – 1 = 3

SWBAT Solve Quadratic Word Problems by Factoring

 Practice: The longer leg of a right triangle is two inches more than twice the length of the shorter leg. The hypotenuse is two inches less than three times the length of the shorter leg. Find the length of thehypotenuse.       x

2x + 23x - 2

(x)2 +(2x + 2)2 = (3x – 2)2x2 + 4x2 + 8x + 4 = 9x2 – 12x + 45x2 + 8x + 4 = 9x2 – 12x + 44x2 – 20x = 04x(x – 5) = 04x = 0 x – 5 = 0 x = 0 x = 5

SWBAT Solve Quadratic Word Problems by Factoring

5. A square is altered so that one dimension is increased by 4, while the other dimension is decreased by 2. The area of the resulting rectangle is 55. Find the area of the original square.   

x

x + 4

x - 2A = 55

(x – 2)(x + 4) = 55x2 + 2x – 63 = 0(x + 9)(x – 7) = 0x = -9 x = 7

72 = 49

SWBAT Solve Quadratic Word Problems by Factoring

Summary1. Define the variables that you want to find with let statements.2. Translate the problem into a mathematical equation. 

3. Get all terms on the same side.

4. Factor

5. Set each factor equal to 0 and solve for x.6. Check your answer(s) with the conditions given in the

problem.

SWBAT Solve Quadratic Word Problems by Factoring

Exit Ticket The square of a number exceeds 5 times the number by 24.  Find the number(s).   

SWBAT Solve Quadratic Word Problems by Factoring