Class 11 Notes I. Radical and Exponential Expressions A. nth Roots and Fractional Powers B. Simplifying II. Systems of Two Linear Equations A. Solving B. Word Problems 1. Speed / Current Problems 2. Mixture Problems 3. Geometry Problems Note to instructor: this class blends two topics that don't really go together -- I don't know why Mercer's syllabus chooses to do this, but we're stuck with it. Generally spend more time on the second subject, because it's harder and more useful. I. Radical and Exponential Expressions Last week in solving quadratic equations, we encountered lots of expressions involving square roots. For half of this class, we'll explore square roots, and things like them, in a bit more detail. A. nth Roots and Fractional Powers Taking the square root of x just means "what number do I have to multiply by itself to get x?" So for example, sqrt(4) = +-2, because 2 . 2 = 4 and (-2) . (-2) = 4. That you already knew. We can generalize this fairly easily to higher roots. For example, suppose we wanted to ask "what number do I have to multiply by itself 3 times to get x?" We call this operation the cube root, and for example cbrt(8) = 2 because 2 . 2 . 2 = 8, or 2^3 = 8 Similarly, we can define the 4th root, 5th root, etc., in completely analogous ways. For example, fourth root(81) = 3, because 3 . 3 . 3 . 3 = 81. At this point we can introduce some new notation which is easier than writing all of these nth root signs. Instead of writing the square root of a number, say 16, as sqrt(16), we'll write it this way: sqrt(16) = 16^(1/2) = 4 The rule is that if you see something to the 1/n power, then that just means take the nth root of it. So for example 16^(1/4) = 2 27^(1/3) = 3 Similarly, we can define expressions where the exponent is any fraction. If you see something like 8^(2/3), the rule is to raise the number to the power in the numerator, and then take the nth root as indicated by the denominator. So for example 8^(2/3) = (8^2)^(1/3) = 64^(1/3) = 4 Of course the order of evaluation doesn't matter, so you can take the root first instead. For example: 81^(3/4) = (81^(1/4))^3 = 3^3 = 27 B. Simplifying We can also recall from earlier in the semester that taking things to negative powers means taking 1 over them, so we can combine these rules naturally. For example, 4^(-1/2) = 1 / 4^(1/2) = 1/2 81^(-3/4) = 1/(81^(1/4))^3 = 1/3^3 = 1/27 The rules for combining powers we've also reviewed before, and extend to fractional powers as well. Thus, when you multiply two bases to different powers, add the powers. When you divide, subtract the powers. For example: x^2 x^(1/2) = x^(2 + 1/2) = x^(5/2) 5^(3/2) / 5^(1/3) = 5^(3/2 - 1/3) = 5^(9/6 - 2/6) = 5^(7/6) When you raise a power to a power, multiply the powers: [x^(2/3)]^(1/4) = x^(2/3 . 1/4) = x^(1/6) So we can write down the complete set of rules for simplifying powers: 1. Raising to a negative exponent: x^(-a) = 1/x^a 2. Multiplying expressions with exponents: x^a x^b = x^(a+b) 3. Dividing expressions with exponents: x^a / x^b = x^(a-b) 4. Raising powers to powers: (x^a)^b = x^(ab) With these, we can simplify complex expressions: Let's try some examples: sqrt(u^8 v^6) / u^(5/2) = u^(8/2) v^(6/2) / u^(5/2) = u^(3/2) v^3 (x^4 y^5 z^6)^(-1/2) = x^(-2) y^(-5/2) z^(-3) = 1 / (x^2 y^(5/2) z^3) II. Systems of Two Linear Equations OK, now to change gears completely. A. Solving To motivate this, let's go back to lines for a minute. It is often convenient to figure out where lines cross. Let's start by considering an example. (on board) Graph y = x - 1 and y = 1/2 x, and find their intersection. OK, this is just graphing lines. How do we do this? (let class answer) (on board) x Eqn 1 y1 Eqn 2 y2 0 y = 0 - 1 -1 y = 1/2 (0) 0 1 y = 1 - 1 0 y = 1/2 (1) 1/2 -1 y = -1 - 1 -2 y = 1/2 (-1) -1/2 2 y = 2 - 1 1 y = 1/2 (2) 1 (draw graphs) As we can clearly see, these two lines both include the point (2,1). We can say that (2,1) solves the two equations simultaneously. We can also see from the graphs that this is the one and only point where the two lines cross. We call equations of this sort, that cross at a single point independent equations Suppose we wanted to find that intersection point without the trouble of drawing graphs. We'd like an algebraic way of finding it. Fortunately, there is such a way. We'll see how this works with an example. Let's solve, or find the intersection of, 2x + y = -1 and x - 2y = 7. We do this using a technique called elimination. The first step is: (on board) 1. Choose a variable to cancel, and multiply one of the equations appropriately. Let's say we want to cancel the y's. In the first equation, we have +y, in the second we have -2y. We want to be able to cancel the y's in the first equation with those in the second, so let's take the first equation and multiply by 2. (on board) 2x + y = -1 2 (2x + y) = 2 (-1) 4x + 2y = -2 Why did I do that? So I can do step 2: (on board) 2. Add or subtract the equations. The two equations we have now are: (on board) 4x + 2y = -2 x - 2y = 7 Let's just add these equations: (on board) 4x + 2y = -2 + x - 2y = 7 -------------- 5x = 5 Now we can see the rationale for multiplying by 2 earlier -- it make the cancellation we see here possible. We're left with one equation in one variable, which we can solve easily enough. (on board) 3. Solve the resulting equation. (on board) 5x = 5 x = 1 The final step is the same as for substitution: (on board) 4. Plug into one of the original equations and solve. Let's do that: (on board) x - 2y = 7 1 - 2y = 7 -1 -1 -2y = 6 y = -3 So the answer is: (on board) x = 1, y = -3, or (1, -3) Let's try another example: (on board) Solve 2x - 3y = -1 and x - 2y = 0. OK, how should we approach this? (let class walk through it) (on board) 2 (x - 2y) = 2 (0) 2x - 4y = 0 (on board) 2x - 3y = -1 -(2x - 4y) = - 0 ---------------- y = -1 (on board) x - 2y = 0 x - 2(-1) = 0 x + 2 = 0 x = -2 (on board) Solution is x = -2, y = -1, or (-2,-1). Let's try one more: x + 3y = 1 and 2x + y = 7 (walk through solving this as time perimits) However, I should add one caution. Suppose we tried to solve these two equations: x - y = 2 3x - 3y = -3 Using elimination, we'll multiply the first equation by 3: 3(x - y = 2) 3x - 3y = 6 If we now try to cancel with the other equation we get 3x - 3y = 6 -(3x - 3y = -3) 0 = 9 Clearly, something went wrong -- the variable cancelled out, and we were left with a false statement. What happened? We'll let's make a quick graph of the two lines to see. (make graph) As you can see, these two lines have exactly the same slope. The lines are parallel, so they never cross. Equations of this form are called inconsistent, and there is no solution. Similarly, suppose we had the two equations x - y = -1 3x - 3y = -3 In this case the variables would cancel and we'd be left with the true statement 0 = 0. This means the two lines are the same, so there are an infinite number of solutions. Equations like this are called dependent. B. Word Problems There are many types of problems that use this information. The project is an example of one. Here we're going to go through some others. 1. Speed/Current Problems A boat takes 4 hours to make a trip downriver, with the current, and 6 hours to make the return trip upriver, against the current. The round trip is 48 miles. Find the speed of the current and the speed of the boat in still water. In a problem like this, we're told something about the amount of time it takes to make a trip, or the length of a trip, and we are asked to find the speed of the object and the speed of the current. In this case we're given the times of each leg of the trip, but we could also have been given distances, or speeds or something like that. In a problem like this, or any problem involving movement, we take an approach like this: (on board) 1. For each leg of the trip, make a table of rate, time, and distance. So let's do that for this problem. (on board) Rate Time Distance Upriver Downriver Next, we will in the values in the table that are given in the problem. (on board) 2. Fill in table values that are given. In this problem, what is given? (let class answer) (on board) Rate Time Distance Upriver 6 24 Downriver 4 24 The next step is perhaps the trickiest: we have to choose our unknowns, and use them to fill in the missing elements of the table. (on board) 3. Choose your unknowns and put them on the table. Generally the best strategy is to choose what you're asked to find as your unknowns. In this type of problem, the unknowns will generally be the speed of the boat, plane, or whatever, when there is no current, and then the speed of the current. Let's use those for our unknowns: (on board) Let b = speed of boat in still water Let c = speed of current Now we need to put these on the table. If b is the speed of the boat in still water, and c is the speed of the current, what is the speed of the boat when it is going upriver, against the current? (let class answer) If the current is 5 miles per hour, and the boat speed is 8 miles per hour, then the speed against the current is just 3, or 8 - 5. So if the speed of the boat is b, and the current is c, the speed against the current is b - c. Similarly, the speed with the current will be b + c. So let's fill those in: (on board) Rate Time Distance Upriver (b-c) 6 24 Downriver (b+c) 4 24 OK, next step: (on board) 4. Use distance = rate x time write equations. This is fairly straightforward. If a car went at 50 miles per hour for 4 hours, how did it go? (let class answer) It went 200 miles, or 50 miles per hour time 4 hours. The idea is the same here: we just multiply the speed by the time to get the distance. If we do that for our problem, we get two equations. (on board) distance = rate . time Upriver: 24 = (b-c) . 6 Downriver: 24 = (b+c) . 4 Now we've reduced the problem to just straight two equations in two variables, and we know how to do those. (on board) 5. Solve the equations. How do we solve these? (let class walk through it) (on board) 6 (b-c) = 24 -> 6b - 6c = 24 4 (b+c) = 24 -> 4c + 4c = 24 (on board) 4 (6b - 6c) = 4 (24) -> 24b - 24c = 96 6 (4b + 4c) = 6 (24) -> + 24b + 24c = 144 --------------- 48b = 240 b = 5 (on board) 6(5) - 6c = 24 30 - 6c = 24 -6c = -6 c = 1 Now we have the answer, and we just need to do the final step: (on board) 6. Write the answer in English. So what is the answer, in English? (let class answer) (on board) The speed of the boat in still water is 5 mph, and the current is 1 mph. That's it. Not all speed/current problems are exactly like that. Some may give you different information. The basic idea of making the table and using distance = rate times time, though, will always work. 2. Mixture Problems In this problem, we are mixing or combining two substances of different values. We might be mixing solutions with different concentrations of a chemical, coins with different values, or anything else similar. All these problems can be solved by fundamentally the same approach. Here's an example. (on board) A jeweler makes two rings, one containing 8g of gold and 4g of silver, the other containing 6g of gold and 6g of silver. Materials for the first ring cost $104, and for the second cost $84. Find the price per gram of gold and silver. For this type of problem we make a table listing each quantity and its value for each object. (on board) 1. Make a table showing each quantity and its value for each object. Here's what I mean: (on board) Gold wgt val/g Silver wgt val/g Total value Ring 1 Ring 2 As before, the next step is to fill in the values that are given. (on board) 2. Fill in values that are given. What information in this table are we given? (let class answer) We are told the weight of gold and silver in each ring, and the total value of each ring. So we can fill those in: (on board) Gold wgt val/g value Silver wgt val/g value Total value Ring 1 8 4 $104 Ring 2 6 6 $84 The next step in the strategy is just like in the previous type of problem. (on board) 3. Choose unknowns and put them on the table. What are the unknowns here? (let class answer) Again, we generally pick the unknowns to be whatever we're asked to find. In this case, it's the value per gram for gold and silver. Let's choose those: (on board) Let g = price per gram for gold Let s = price per gram for silver Now we need to fill in the table. How should we fill in the table here? (let class answer) We know that the price per gram for gold is g, so that just goes in the "value per gram column", and similarly for silver. The value of the gold in the ring is just the number of grams times the price per gram, so we can fill that in too. (on board) Gold wgt val/g value Silver wgt val/g value Total value Ring 1 8 g 8g 4 s 4s $104 Ring 2 6 g 6g 6 s 6s $84 Now we can go to the next step. (on board) 4. Write equations using the total values. What equations can we write here? (let class answer) We know that the value of the gold in ring 1 is 8g, and the silver is 4s, so the total value is just 8g + 4s. A similar idea applies for ring 2. Thus, we can write (on board) 8g + 4s = 104 6g + 6s = 84 Now we're down to pure symbol manipulation again. (on board) 5. Solve the equations. How do we solve these? (let class walk through it) (on board) 3 (8g + 4s) = 3 (104) -> 24g + 12s = 312 2 (6g + 6s) = 2 ( 84) -> - (12g + 12s) = - 168 ------------------- 12g = 144 g = 12 8 (12) + 4s = 104 96 + 4s = 104 4s = 8 s = 2 Then we just have to do the last step: (on board) 6 Write the answer in English. What's the answer in English? (let class answer) (on board) Gold costs $12/g, silver costs $2/g. 3. Geometry Problems The final type of problem I want to discuss today is one that involves some geometry. Generally, this type of problem involves some sort of shape, and we want to figure things out about that shape. These problems are very similar to number problems, except that some of the sentences we have to translate into algebra are in terms of geometric concepts. Here's an example. (on board) A farmer is fencing a rectangular pasture. He wants the enclosed area to have a total perimeter of 220 meters. The pasture's length is 10 meters less than twice its width. Find the dimensions of the pasture. The first step in a problem like this, or any geometry problem, is always (on board) 1. Draw a picture. In this case we have a rectangular pasture, so let's draw that. (draw rectangle on board) The next step is, as one might expect (on board) 2. Choose variables and label the drawing with them. What do you think we should pick? In this sort of problem, we generally choose as variables the things we're being asked to solve for. In this case we want the dimensions of the field, so let's use those: (on board) Let l = length w = width Once we've chose our variables, we label them on the drawing. (draw labels on the drawing) Now we proceed as for a normal number problem: (on board) 3. Translate each of statement into an equation. The first statement is that the pasture will have a perimeter of 220 meters. What equation can we write from this? (let class answer) (on board) 2l + 2w = 220 OK, now let's try the next sentence, that the length is 10 meters less than twice the width. What equation can we write for that? (let class answer) l = 2w - 10 Now we solve. (on board) 4. Solve the equations. How do we solve these? (let class walk through it) (on board) l = 2w - 10 2l + 2w = 220 -> 2(2w - 10) + 2w = 220 4w - 20 + 2w = 220 6w = 240 w = 40 l = 2w - 10 = 2(40) - 10 = 70 And then the final step. (on board) 5. Write your answer in English. (on board) The pasture is 40 meters by 70 meters.