Class 9 Notes I. Rational Equations A. Solving B. Word Problems II. Complex Numbers A. Square Root of -1 B. Addition / subtraction C. Multiplication D. Division I. Rational Equations A. Solving We've spent the last week working with rational expressions, and this week we're going to learn how to solve equations with them. Let's take an example equation that involves rational expressions of the type we dealt with last week: 3 / (x + 1) + 3 / (x - 1) = 4 These types of equations are fairly common in problems involving rates, as we'll see shortly. When we see something like this, with the variables in the denominator, our goal is to turn it into an equation we know how to solve by getting all the variables out of the denominator. Here's how we proceed: 1. Find the LCD of all the terms. This is exactly what we practiced last week when we added and subtracted rational expressions. In this case, what's the LCD? (let class answer) It's (x+1)(x-1). Next step: 2. Multiply both sides of the equation by the LCD and simplify. We multiply everything by the LCD. This gives (x+1)(x-1)[3/(x+1) + 3/(x-1)] = (x+1)(x-1)(4) 3x-3 + 3x+3 = 4x^2 - 4 6x = 4x^2 - 4 3. Solve the resulting equation. The equation we've gotten to is just a quadratic, which we know how to solve. 4x^2 - 6x - 4 = 0 2x^2 - 3x - 2 = 0 (2x + 1)(x - 2) = 0 x = -1/2, 2 So we have a solution, but we need to do one more step. 4. Check the solution in the original equation. We just write down the original equation, substitute in our solutions, and make sure it works. 3/(x + 1) + 3/(x - 1) = 4 Trying x = -1/2: 3/(-1/2+1) + 3/(-1/2-1) = 3/(1/2) + 3/(-3/2) = 6 - 2 = 4 So this checks. No trying x = 2: 3/(2+1) + 3/(2-1) = 3/3 + 3/1 = 3 + 1 = 4 So this checks too, and we're done. Why did we bother checking? To see why, let's try another example: x/(x+3) + 2/(x-3) = (2x+6) / (x^2-9) Let's do this step by step. First we find the LCD. To do this, we have to factor all the denominators. The terms x+3 and x-3 are already factored, so we factor x^2 - 9: x^2 - 9 = (x+3)(x-3) So the LCD is (x+3)(x-3). Now we multiply by it: (x+3)(x-3) [x/(x+3) + 2/(x-3)] = (x+3)(x-3)[(2x+6) / (x^2-9)] x(x-3) + 2(x+3) = 2x + 6 x^2 - 3x + 2x + 6 = 2x + 6 x^2 - 3x = 0 Now we solve. Fortunately, this equation is easy to solve: x(x-3) = 0 x = 0, 3 Now let's check. First we set x = 0: x/(x+3) + 2/(x-3) = (2x+6) / (x^2-9) 0/(0+3) + 2/(0-3) = (2(0) + 6) / (0^2 - 9) 0 + 2/(-3) = 6/(-9) -2/3 = -2/3 So this checks. Now check x = 3: x/(x+3) + 2/(x-3) = (2x+6) / (x^2-9) 0/(3+3) + 2/(3-3) = (2(3) + 6) / (3^2 - 9) 0/6 + 2/0 = 12/0 Here we have a problem. Two of these terms are undefined, since they have zero in the denominator. That's why we need to check. Multiplying by the LCD, as we have done, can introduce spurious solutions. In this case, only x = 0 is a valid solution. As a side note: what went wrong that made a spurious solution appear? The reason is that we multiplied by (x-3)(x+3), and when x = 3 this means that we multiplied by 0. Of couse multiplying by 0 is not allowed, which is why we got an invalid solution. B. Word Problems One major place that rational equations come up is in rate problems, where we're told something about how long a combined process takes and asked to figure out how long each part would take by itself. Here's an example: Andy and Bob take 2 hours to mow a lawn together. Working alone, it would take Bob 3 hours more than Andy. How long would each one take to mow the lawn alone? This is what is known as a "rate of work problem". Here's how we approach it: (on board) 1. Make a table showing rate of work, time, and amount done for each participant. (on board) Rate of Work Time Amount Done Andy Bob Next, we fill in what we know. (on board) 2. Fill in the known quantities. (on board) Rate of Work Time Amount Done Andy 2 Bob 2 The third step is choosing a variable. In a problem like this, we choose the time one of the two would need to complete the task by himself as the variable. We write the time the other one would need in terms of this. (on board) 3. Choose the time needed by one of the participants alone as the variable, and write the time needed by the other participants in terms of it. In this case, we know Bob would need 3 hours more than Any if he were working alone. (on board) Let t = time needed by Andy t + 3 = time needed by Bob The next step is the key to the problem. (on board) 4. The rate of work for each participant is 1 divide by that participant's time. Fill this in. Here's the idea: suppose we were to find that it takes Andy 2 hours to mow the lawn by himself. In that case, we could say that he mows 1/2 of the lawn per hour. If Bob took 5 hours, we could mow 1/5 of the lawn per hour. Thus, the rate is just 1 divided by the time. So let's fill that in. (on board) Rate of Work Time Amount Done Andy 1 / t 2 Bob 1 / (t + 3) 2 Now we multiply rate by time, as we have in earlier problems. (on board) 5. Multiply rate times time to get the amount done. (on board) Rate of Work Time Amount Done Andy 1 / t 2 2 / t Bob 1 / (t + 3) 2 2 / (t + 3) The total amount done is 1 lawn, so we can use this to write an equation. (on board) 6. The total amount done is 1. Write an equation using that. (on board) 2 / t + 2 / (t+3) = 1 Now we're down to algebra. 7. Solve the equation. This is just like the equations we've been solving. The LCD is t(t+3), so we multiply by it: (on board) t(t+3) (2/t + 2/(t+3)) = 1 . t(t+3) 2(t+3) + 2t = t(t+3) Now we solve. Someone walk through solving this. (let class walk through it) (on board) 2t + 6 + 2t = t^2 + 3t t^2 - t - 6 = 0 (t - 3)(t + 2) = 0 t = -2, 3 We have two answers here, but only one of them makes sense. After all, it's pretty hard to take -2 hours to mow a lawn. Thus, we have the final step: (on board) 8. Choose the solution that makes sense, and write the answer in English. (on board) It would Andy 3 hours by himself. It would take Bob 6 hours. II. Complex Numbers Now we're going to change gears completely and talk about a peculiar but very important idea: complex numbers. A. Square Root of -1 Explain idea of expanding number systems, definition of i Show sqrt(-1) = i, i^2 = -1 Show how to evaluate square root of any negative number in terms of i Example: sqrt(-5) = sqrt(-1) * sqrt(5) = i * sqrt(5) Every complex number can be written as a real part plus an imaginary part. Example: 3 + 2i. B. Addition / subtraction Go over addition and subtraction of complex numbers Example: 2 + 3i + (4 - 2i) = 6 + i 2 + 3i - (4 - 2i) = -2 + 5i C. Multiplication Explain how to multiply imaginary and complex numbers using FOIL Example: (2 + 3i) * (4 - 2i) = 8 - 4i + 12i - 6i^2 = 8 + 8i - 6(-1) = 14 + 8i Another example: sqrt(-2) * sqrt(-10) = i * sqrt(2) * i * sqrt(10) = i^2 * sqrt(20) = -sqrt(20) D. Division Division by pure imaginaries Example: (2 + 3i) / (2i) = (2 + 3i) * i / (2i * i) = (-3 + 2i) / (-2) = 3/2 - i Definition of conjugates and division of complex numbers: Conjugate of 2 + 3i = (2 + 3i)* = 2 - 3i Example: (2 + 3i) / (4 - 2i) = (2 + 3i) * (4 + 2i) / [(4 - 2i) * (4 + 2i)] = (4 + 4i + 12i + 6i^2) / (16 + 8i - 8i - 4i^2) = (-2 + 16i) / (20) = -1/10 + 4/5 i