Study Guide for Skills Module 3: An Approximate Method for Solving Quadratic Equations

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1 Chemistry 1020, Skills Module 3 Name Study Guide for Skills Module 3: An Approximate Method for Solving Quadratic Equations Reading Assignment: The module contains a review of the mathematics needed to work the problems as indicated. Guide for Your Lecturer: 1. Solving a Quadratic Equation by the Exact Method 2. Solving a Quadratic Equation by the Approximate Method Homework Solving A Quadratic Equation By the Exact Method In mathematics courses you have learned that, given the quadratic equation ax 2 + bx + c = 0, the values of x which satisfy this equation can be determined from x = -b + b2-4ac 2a and x = -b - b2-4ac 2a 1. Solve the quadratic equation by the exact method. S. x 2 / ( x) = 1.0 * 10-3 x x = 1.0*10-3 x 2 = 1.00*10-3 ( x) x 2 = 4.00* *10-3 x x *10-3 x *10-4 = 0 This equation is in the form ax 2 + bx + c = 0 where a = 1, b = 1.00*10-3, c = -4.00*10-4 solve using x = -b ± b2-4ac 2a x = -1.00*10-3 ± (1.00*10-3 ) 2-4*1*(-4.00*10-4 ) 2*1-1.00*10 x = -3 ± 1.601* = 2 which rounds to 2.0*10-2 or -2.0* *10-3 ± 4.00*10-2 = -1.00*10-3 ± 1.00* * = 1.95*10-2 or -2.05*10-2 = A. x 2 / (0.50-2x) = 4.5 * 10-8 Xavier University of Louisiana 111

2 Chemistry 1020, Skills Module 3 Solving A Quadratic Equation By the Exact Method (continued) 1. Solve the quadratic equation by the exact method. B. x 2 / ( x) = 6.2 * 10-4 C. x 2 / ( x) = 1.0 * Xavier University of Louisiana

3 Chemistry 1020, Skills Module 3 Name Solving A Quadratic Equation By the Exact Method (continued) 1. D. x 2 / ( x) = 1.3* 10-6 E. x 2 /( x) = 6.7*10-6 Solving A Quadratic Equation By An Approximate Method The quadratic equations chemists most often encounter do not yield exact results even if solved by the exact method above because the constants in the equation are measured quantities which contain uncertainty. Therefore, it doesn't make any sense to use the "exact method" above if a faster, less tedious approximate method will give an answer which is within the limits of the inherent error in the measured constants. The approximate method chemists most often use to solve quadratic equations is a kind of a successive approximation. The following is an outline of the procedure to be used in solving the type of quadratic equations you will encounter in this chemistry course. The typical quadratic equation which you will encounter in chemistry usually arises in the form x 2 /(d - x) = e where d and e are constants. Therefore we will start the procedure from that point rather than from the "standard" form of ax 2 + bx + c = 0. If you look closely at a quadratic equation it is immediately apparent that the difficulty in solving the equation comes from the x term. If that term could be eliminated, the equation would reduce to the simple one x 2 = d * e which can be solved by just taking a square root. We now use this observation to proceed as follows: Xavier University of Louisiana 113

4 Chemistry 1020, Skills Module 3 Step 1: Problem-Solving Procedure Comments about Procedure Given: x 2 d - x = e Assume that d >>x so that d - x = d This is the same as saying that that x = 0 in the for all practical purposes. (d - x) term. The assumption means that we don't know whether or not the answer we will get is okay The equation now becomes unless we test it. x1 = + d*e which can easily be solved to get x1. Now check to see if the answer is This is the same as checking to see if the aproxi- "close enough" to the exact answer by mate answer is within 5% of the exact answer. looking at x1 /d to see if it is less than Fortunately, most (about 70%) of the quadratic or equal to equations you will see in chemistry can be solved by the approximate method in only one step. What to do if the approximate method doesn't work in the first step (i.e. if the check isn't less than If the check for the first approximation isn't less than or equal to 0.05, look to see how large it is. If it is between 0.05 and 0.25, it is generally faster to go on to a second and third (if necessary) step in the approximate method rather than to go back to the exact method. If the check for x is greater than 0.25, the approximate method probably won't work, so go back to the exact method. Step 2: Assume that x in d - x is equal to x1 You should note that even though x1 wasn't good enough to use permanently, it is good enough to use so that the equation now becomes in getting a second, better approximation to the answer, x2. x2 = + (d - x1)*e Now check the answer by looking to You get percent difference between two numbers see if the %-difference between x1 by looking to see if and x2 is less than or equal to 5%. If so, x2 is good enough to use. If is less than not, go on to Step 3. ABS VALUE (x1 - x2) smaller of x1 or x2 Step 3: Assume that x in d - x is equal to x2 You should note that even though x2 wasn't good enough to use permanently, it is good enough to use so that the equation now becomes in getting a second, better approximation to the answer, x3. x3 = + (d - x2)*e Now check the answer by looking to You get percent difference between two numbers see if the %-difference between x2 by looking at and x3 is less than or equal to 5%. If so, x3 is good enough to use. If is less than not, go on to Step 4 ABS VALUE (x2 - x3) smaller of x2 or x3 Pictorially, this method might be represented as a number of steps each of which (hopefully) give a value closer to the true value than the last. For example, (continued) l l l l x2 true value x3 x1 114 Xavier University of Louisiana

5 Chemistry 1020, Skills Module 3 Name Points to note: 1. This method won't always work! For example, if x 1 = 0 or if x 1 = x 2, there is something weird happening mathematically that prevents it from working. This is probably because the initial guess of x was too far from the value for the "stepping-in" to work out. If you observe either of these or if the successive values of x are not getting closer together, first check your math and if you can find no mistakes them go back to the exact method. 2. In the above discussion, we have only used the positive values when taking the square root. Chemists often do this because the negative answers make no chemical sense. For example, if x were a concentration, a negative value makes no sense--it is not possible to have less than no solute in a solution. DON'T PANIC--THIS METHOD IS FASTER (when it works) AND LEADS TO FEWER MATH MISTAKES THAN DOES THE EXACT METHOD--EVEN IF YOU USE CALCULATORS. The time you save when it works more than makes up for that you lose when it doesn't. 2. Solve the indicated problem by the approximate method. S. x 2 / ( x) = 3.8 * 10-6 Step 1: Assume 0.40 >> x so that 0.40-x 0.40 and the equation becomes x = 3.8*10-6 So x1 2 = (0.40)*(3.8*10-6 ) x1 2 = 1.52*10-6 x1 = 1.52*10-6 = 1.23*10-3 x1 Now check 0.40 = 1.23* = < 0.05 so x1 is ok to use. R Therefore, the solution to the equation is 1.23*10-3 = 1.2*10-3 A. x 2 / ( x) = 9.0 * 10-5 Xavier University of Louisiana 115

6 Chemistry 1020, Skills Module 3 Solving A Quadratic Equation By An Approximate Method (continued) 2. B. x 2 / ( x) = 5.0 * 10-6 C. x 2 / ( x) = 7.5 * 10-6 D. x 2 / ( x) = 4.0 * Xavier University of Louisiana

7 Chemistry 1020, Skills Module 3 Name Solving A Quadratic Equation By An Approximate Method (continued) 2. E. x 2 / ( x) = 5.2 * 10-6 F. x 2 / ( x) = 2.7 * 10-6 (over) Xavier University of Louisiana 117

8 Chemistry 1020, Skills Module 3 Lagniappe: A sample problem in which the check for the first step of the approximate method doesn t work. You will need it for a challenge question later in the course. Solve for x in x 2 / ( x) = 2.0 * 10-3 x 2 / ( x) = 2.0 * 10-3 Step 1: Assume 0.20 >> x so that the equation becomes x = 2.0*10-3 So x1 2 = (0.20)*(2.0*10-3 ) x1 2 = 0.40*10-3 x1 = 0.40*10-3 = 2.0*10-2 x1 Check: 0.20 = 2.0* = 0.10 > 0.05 so x1 is NOT close enough to use. However, it IS close enough (<0.25) to continue with the approximate method. Step 2: Step 3: Assume x = x1 in 0.20-x so that the equation becomes x = 2.0*10-3 So x2 2 = (0.180)*(2.0*10-3 ) x2 2 = 0.360*10-3 x2 = 0.360*10-3 = 1.90*10-2 x2-x1 Check: smaller of X1 of X2 = 1.90* * *10-2 = > 0.05 so x2 also is not close enough to use. Therefore, continue to step 3. Assume x = x2 in 0.20-x so that the equation becomes x = 2.0*10-3 So x3 2 = (0.181)*(2.0*10-3 ) x3 2 = 0.362*10-3 x3 = 0.362*10-3 = 1.90*10-2 x3-x2 Check: smaller of X2 of X3 = 1.90* * *10-2 = 0 < 0.05 so x3 is ok. R Therefore, x = 1.90*10-2 = 1.9*10-2 Revised: JWC 1997; Etim Eduok 2000; JWC 2001; JWC 2002;RI Xavier University of Louisiana