Math Misconceptions and Considerations HSA-REI.C.7
Look closely at errors in students’ work (formative assessment) to help you reflect and make instructional decisions to suit all students’ needs.
One size doesn’t fit all! Students approach the task mechanically and algorithmically instead of conceptually, and they tend to commit to a single method instead of choosing the best strategy for each situation. Misconception: Given the following system of linear and quadratic equations, the student chooses to solve it graphically. 2
y=x +5x−9 y=5x
What to do: If the student were to solve this system algebraically, they would arrive at the solution for x much quicker. x 2 +5x−9=5x 2 x −9=0 x 2=9 x=Âą 3 Therefore, the solutions are (3, 15) and (-3, -15). As a teacher, the discussion of what method is advantageous to different systems should occur. Giving students opportunities to choose the appropriate method for solving the system will deepen their understanding of the concept.
What does it all mean? When solving systems of equations, students often lose sight of what the solution to such a system means. Planning and following correct solving procedures can be problematic when students do not know what the solution re-presents or why the procedure works. Checking answers by verifying the validity of solutions can be difficult for the same reasons. Misconception: When presented with the given set of equations for a problem, students immediately focus on solving the system and not interpreting the answers. The math whiz kids working for the racetrack owners have estimated revenues to be modeled by x2 + 10x – 78 = y while the costs are y = 17x (where y is in hundreds of dollars and x is in hundreds of tickets sold). How many tickets must be sold for the racetrack guys just to break even?
At this point in the problem, many students think that they are done with the problem. What to do: Using the same example from above, we need to have students frame their answers in the context of the problem. The variable x represents the number of tickets sold, in hundreds. Thus, the solutions x = -6 and x = 13 in this problem represents selling -600 tickets and 1300 tickets. Students should be able to eliminate -600 as an answer because you can’t sell a negative number of tickets.