Algebra 1 - connection to physics

1. Solving Equations by Factoring

2. Zero – Product Property For all real numbers a and b, ab = 0 if and only if a = 0 or b = 0

3. Example 1 Solve (x + 2)(x + 5) = 0 By the Zero - Product Property, one of the factors on the left must equal zero. Therefore either (x + 2) = 0 or (x + 5) = 0

4. Example 1 Continued Either (x + 2) = 0 OR (x + 5) = 0 x + 2 – 2 = 0 – 2 x = – 2 x + 5 – 5 = 0 – 5 x = – 5 The solution set is { -2, -5} Partner Try: Solve (y + 5)( y – 7) = 0 Either (y + 5) = 0 OR (y – 7) = 0 y + 5 – 5 = 0 – 5 y + 5 – 7 = 0 + 7 y = – 5 y = 7 The solution set is {-5, 7}

5. Example 2 Solve 5n(n – 3)(n – 4) = 0 Either 5n = 0 or (n – 3) = 0 or (n – 4) = 0 n = 0 (By the zero-product rule) n – 3 + 3= 0 + 3 n – 3 + 4= 0 + 4 n = 3 n = 4 The solution set is {0, 3, 4} Partner Try: Solve 3x(2x + 1)(2x + 5) Either 3x = 0 or (2x + 1) = 0 or (2x + 5) = 0 2x + 1 – 1 = 0 – 1 x = 0 (By the zero-product rule) 2x + 1 – 5 = 0 – 5 2x = – 1 2x = – 5 x = – 1 2 x = – 5 2 The solution set is {0, –1 , – 5} 2 2

6. Example 3 Solve 2x2 + 5x = 12 Step 1: Transform the equation into standard form. 2x2 + 5x – 12 = 0 Step 2: Factor the left side. (2x – 3)(x + 4) = 0 Step 3: Set each factor equal to zero & solve. 2x – 3 = 0 x + 4 = 0 2x – 3 + 3 = 0 + 3 2x = 3 x = 3 2 x + 4 – 4 = 0 – 4 x = – 4 The solution set is {3 , – 4} 2

7. Partner Try Solve y2 – 3y + 2 = 0 Step 1: The equation is already in standard form. Step 2: Factor the left side. (y – 2)(y – 1) = 0 Step 3: Set each factor equal to zero and solve. y – 2 = 0 y – 1 = 0 y – 2 + 2 = 0 +2y – 1 + 1 = 0 + 1 y = 2 y = 1 The solution set is {1, 2}

8. Physics Problems In Physics, often times we need to use factoring to answer questions about time, height and speed. Let’s use the variables h for height,rfor rate of speed, and tfor time. Physicists have learned through experimentation that the height, rate of speed, and time of an object are related. They describe this relation using a mathematical formula. h = rt -16t2. We can use this formula to find height, rate of speed, or time if we know the other two variables.

9. Physics Problem #1 A ball is kicked upward with an initial speed of 20 m/s. When is it 6 m high? Step 1: Draw a picture 6m 6m 6m 6m Possibility 1: 2 solutions Possibility 2: 1 solution Possibility 3: 0 solutions

10. Physics Problem #1 Step 2: Identify the variables. h = 6 mr = 20 m/st = time Step 3: Write an equation using the physics formula h = rt – 16t2. 6 =20(t) – 16(t)2 Step 4: Simplify & Solve 16(t)2 – 20(t) + 6 = 20(t) – 20(t) – 16(t)2+ 16(t)2 16(t)2 – 20(t) + 6 = 0 (4t – 3)(4t – 2 ) = 0 4t – 3 = 0 or 4t – 2 = 0 4t – 3 + 3 = 0 + 3 4t – 2 + 2 = 0 + 2 4t = 3 4t = 2 t = 3 = .75 t = 2 = .5 4 4

11. Physics Problem #1 Step 5: Check all possible solutions. If t = .75 [4(.75) – 3][4(.75) – 2] = [3 – 3][3 – 2] = (0)(1) = 0 If t = .5 [4(.5) – 3][4(.5) – 2] = [– 1][2 – 2] = (-1)(0) = 0 Therefore, ball is at 6 meters at .5 seconds and .75 seconds. ✓ ✓ 6m 6m

12. Partner Try Katie hit a softball upward with an initial speed of 120 ft/s. How much later did the catcher catch it? Step 1: Draw a picture Step 2: Identify the variables h = height when the catcher catches the ball = 0 r = rate of speed = 120 ft/s t = time catcher caught it = t Step 3: Write an equation using the physics formula h = rt -16(t)2 so 0 = 120(t) – 16(t)2

13. Partner Try Step 4: Simplify and Solve 0 = 120t – 16t2 0 = 16t(7.5 – t) Factoring out 16t from both terms. Either 16t = 0 or 7.5 – t = 0 t = 0 or 7.5 – t+t = 0 + t t = 7.5

14. Partner Try Step 5: Check all possible solutions. If t = 0, that means the catcher would catch the ball the moment it is hit, which is not possible. Therefore, t cannot equal 0. If t = 7.5, 0 =? 120(7.5) – 16(7.5)2 0 = 900 – 900 ✓ Therefore, the catcher will catch the ball after 7.5 seconds.